def addNum(self, num):
     """
     Adds a num into the data structure.
     :type num: int
     :rtype: void
     """
     if len(self.left) > len(self.right):
         if num > self.median:
             heapq.heappush(self.right, num)
         else:
             heapq.heappush(self.right, -heapq.heappop(self.left))
             heapq.heappush(self.left, -num)
         self.median = (-self.left[0] + self.right[0]) / 2.0
     elif len(self.left) == len(self.right):
         if num > self.median:
             heapq.heappush(self.right, num)
             self.median = self.right[0]
         else:
             heapq.heappush(self.left, -num)
             self.median = -self.left[0]
     else:
         if num < self.median:
             heapq.heappush(self.left, -num)
         else:
             heapq.heappush(self.left, -heapq.heappop(self.right))
             heapq.heappush(self.right, num)
         self.median = (-self.left[0] + self.right[0]) / 2.0
Пример #2
0
    def _get_server(self):
        """
        Get server to use for request.
        Also process inactive server list, re-add them after given interval.
        """
        with self._lock:
            inactive_server_count = len(self._inactive_servers)
            for i in range(inactive_server_count):
                try:
                    ts, server, message = heapq.heappop(self._inactive_servers)
                except IndexError:
                    pass
                else:
                    if (ts + self.retry_interval) > time():
                        # Not yet, put it back
                        heapq.heappush(self._inactive_servers,
                                       (ts, server, message))
                    else:
                        self._active_servers.append(server)
                        logger.warn("Restored server %s into active pool",
                                    server)

            # if none is old enough, use oldest
            if not self._active_servers:
                ts, server, message = heapq.heappop(self._inactive_servers)
                self._active_servers.append(server)
                logger.info("Restored server %s into active pool", server)

            server = self._active_servers[0]
            self._roundrobin()

            return server
Пример #3
0
    def _vsids_calculate(self):
        """
            VSIDS Heuristic Calculation

        Examples
        ========

        >>> from sympy.logic.algorithms.dpll2 import SATSolver
        >>> l = SATSolver([set([2, -3]), set([1]), set([3, -3]), set([2, -2]),
        ... set([3, -2])], set([1, 2, 3]), set([]))

        >>> l.lit_heap
        [(-2.0, -3), (-2.0, 2), (-2.0, -2), (0.0, 1), (-2.0, 3), (0.0, -1)]

        >>> l._vsids_calculate()
        -3

        >>> l.lit_heap
        [(-2.0, -2), (-2.0, 2), (0.0, -1), (0.0, 1), (-2.0, 3)]
        """
        if len(self.lit_heap) == 0:
            return 0

        # Clean out the front of the heap as long the variables are set
        while self.variable_set[abs(self.lit_heap[0][1])]:
            heappop(self.lit_heap)
            if len(self.lit_heap) == 0:
                return 0

        return heappop(self.lit_heap)[1]
Пример #4
0
def huff_encode(text):
	freq = defaultdict(int)
	for s in text:
		freq[s] += 1

	tree = [ [f, [s, ""]] for s,f in freq.items() ]
	heapify(tree)

	while len(tree) > 1:
		l = heappop(tree)
		h = heappop(tree)
		
		for n in l[1:]:
			n[1] = '0' + n[1]
		for n in h[1:]:
			n[1] = '1' + n[1]

		heappush(tree, [l[0] + h[0]] + l[1:] + h[1:])

	root = heappop(tree)[1:]
	codes = dict([(s, "0b"+c) for s,c in root])
	
	# Header
	enc = BitArray()
	for s,c in root:
		enc += BitArray(bytes=s)
		enc += BitArray(uint=len(c), length=8)
		enc += BitArray("0b"+c)
	enc.prepend(BitArray(uint=len(root), length=8))
			
	for s in text:
		enc += BitArray(codes[s])

	return enc
Пример #5
0
    def create_binary_tree(self):
        """
        Create a binary Huffman tree using stored vocabulary word counts. Frequent words
        will have shorter binary codes. Called internally from `build_vocab()`.

        """
        logger.info("constructing a huffman tree from %i words" % len(self.vocab))

        # build the huffman tree
        heap = self.vocab.values()
        heapq.heapify(heap)
        for i in xrange(len(self.vocab) - 1):
            min1, min2 = heapq.heappop(heap), heapq.heappop(heap)
            heapq.heappush(heap, Vocab(count=min1.count + min2.count, index=i + len(self.vocab), left=min1, right=min2))

        # recurse over the tree, assigning a binary code to each vocabulary word
        if heap:
            max_depth, stack = 0, [(heap[0], [], [])]
            while stack:
                node, codes, points = stack.pop()
                if node.index < len(self.vocab):
                    # leaf node => store its path from the root
                    node.code, node.point = codes, points
                    max_depth = max(len(codes), max_depth)
                else:
                    # inner node => continue recursion
                    points = array(list(points) + [node.index - len(self.vocab)], dtype=uint32)
                    stack.append((node.left, array(list(codes) + [0], dtype=uint8), points))
                    stack.append((node.right, array(list(codes) + [1], dtype=uint8), points))

            logger.info("built huffman tree with maximum node depth %i" % max_depth)
def skyline(buildings):
    if not buildings:
        return []
    points = []
    for l, r, h in buildings:
        points.append([l, -h])
        points.append([r, h])
    # sort the points to ensure that all points are sorted from left to right
    points.sort()
    result = []
    queue = [0]
    prev = 0
    for e in points:
        if e[1] < 0:
            # whenever we see start point of the building
            # we put the height of the building into the heapq
            heapq.heappush(queue, e[1])
        else:
            # if we arrive at the end point of the building
            # we remove the height of the building from the heapq
            if queue[0] == -e[1]:
                heapq.heappop(queue)
            else:
                queue.remove(-e[1])
                heapq.heapify(queue)
        curr = queue[0]
        if curr != prev:
            # if the tallest building in the heapq changes
            # then we need to put that into the final result
            result.append([e[0], -curr])
            prev = curr
    return result
	def test_buscar_el_mayor_y_menor_valor_de_una_lista(self):
		import heapq
		nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]

		tres_mayores = heapq.nlargest(3,nums) 
		self.assertEqual(tres_mayores,[42,37,23])
		tres_menores = heapq.nsmallest(3,nums)
		self.assertEqual(tres_menores,[-4,1,2])

		mayor = max(nums)
		menor = min(nums)
		self.assertEqual(mayor ,42)
		self.assertEqual(menor ,-4)

		heap = list(nums)
		#Pone el primer elemento en la posicion 0
		heapq.heapify(heap)
		self.assertEqual(heap,[-4, 2, 1, 23, 7, 2, 18, 23, 42, 37, 8])

		#heappop saca el primer elemento, y lo sustituye por el menor de la lista
		# asi heappop siempre sacara el elemento mas pequeño
		menor =heapq.heappop(heap)
		self.assertEqual(menor,-4)
		menor =heapq.heappop(heap)
		self.assertEqual(menor, 1)	
Пример #8
0
 def _build_tree(self, a):
     heapq.heapify(a)
     while len(a) > 1:
         left = heapq.heappop(a)
         right = heapq.heappop(a)
         heapq.heappush(a, (left[0] + right[0], left, right))
     return a[0]
Пример #9
0
    def get_next_task(self):
        """get the next task if there's one that should be processed,
        and return how long it will be until the next one should be
        processed."""
        if _debug: TaskManager._debug("get_next_task")

        # get the time
        now = _time()

        task = None
        delta = None

        if self.tasks:
            # look at the first task
            when, nxttask = self.tasks[0]
            if when <= now:
                # pull it off the list and mark that it's no longer scheduled
                heappop(self.tasks)
                task = nxttask
                task.isScheduled = False

                if self.tasks:
                    when, nxttask = self.tasks[0]
                    # peek at the next task, return how long to wait
                    delta = max(when - now, 0.0)
            else:
                delta = when - now

        # return the task to run and how long to wait for the next one
        return (task, delta)
    def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
        s = sneg = neg = 0
        absmin = math.inf
        negpq = []
        for a in A:
            absmin = min(absmin, abs(a))
            if a >= 0:
                s += a
            else:
                heapq.heappush(negpq, a)
                neg += 1
                sneg += a

        if K < neg:
            while K:
                s -= heapq.heappop(negpq)
                K -= 1
            while negpq:
                s += heapq.heappop(negpq)
        else:
            s -= sneg
            if (K - neg) & 1:
                s -= absmin * 2

        return s
Пример #11
0
	def nextTime(self):
		while True:
			(t, k, o) = self._heap[0]
			if k in self._dict:
				break
			heapq.heappop(self._heap)
		return t
Пример #12
0
    def getImminent(self, time):
        """
        Returns a list of all models that transition at the provided time, with a specified epsilon deviation allowed.

        :param time: timestamp to check for models

        .. warning:: For efficiency, this method only checks the **first** elements, so trying to invoke this function with a timestamp higher than the value provided with the *readFirst* method, will **always** return an empty set.
        """
        #assert debug("Asking all imminent models")
        imm_children = []
        t, age = time
        try:
            # Age must be exactly the same
            first = self.heap[0]
            while (abs(first[0][0] - t) < self.epsilon) and (first[0][1] == age):
                # Check if the found event is actually still active
                if(first[2]):
                    # Active, so event is imminent
                    imm_children.append(first[3])
                    first[2] = False
                else:
                    # Wasn't active, but we will have to pop this to get the next
                    # So we can lower the number of invalids
                    self.invalids -= 1

                # Advance the while loop
                heappop(self.heap)
                first = self.heap[0]
        except IndexError:
            pass
        return imm_children
Пример #13
0
    def generateAckPacket(self, packet):
        '''
            Generate ACK packet
            1) If packet.pck_id == self.lastOrderedPacketID+1, the packet is the packet to be expected
               else, push the pck_id to the out of order queue
            2) Generate ACK packet with the lastOrderedPacketID
            3) set the same timestamp for the ACK Packet as the original Data Packet
            
            Args:
                packet: the Data Packet received
        '''
        if packet.pck_id <= self.lastOrderedPacketID:
            pass
        else:
            if packet.pck_id == self.lastOrderedPacketID+1:
                self.lastOrderedPacketID += 1
                
                while self.outOfOrderPackets and self.outOfOrderPackets[0] == self.lastOrderedPacketID+1:
                    heapq.heappop(self.outOfOrderPackets)
                    self.lastOrderedPacketID +=1
            else:
                heapq.heappush(self.outOfOrderPackets, packet.pck_id)

        ack_packet = DataPacket(packet.destination, packet.source, self, self.engine.getCurrentTime(), ACK_PACKET_SIZE, 
                                True, self.lastOrderedPacketID + 1)
        ack_packet.setOriginalPacketTimestamp(packet.timestamp)

        return ack_packet
Пример #14
0
 def __init__(self, search_strategy):
     if search_strategy == _DEPTH_FIRST:
         #use stack for OPEN set (last in---most recent successor
         #added---is first out)
         self.open = []
         self.insert = self.open.append
         self.extract = self.open.pop
     elif search_strategy == _BREADTH_FIRST:
         #use queue for OPEN (first in---earliest node not yet
         #expanded---is first out)
         self.open = deque()
         self.insert = self.open.append
         self.extract = self.open.popleft
     elif search_strategy == _BEST_FIRST:
         #use priority queue for OPEN (first out is node with
         #lowest hval)
         self.open = []
         #set node less than function to compare hvals only
         sNode.lt_type = _H
         self.insert = lambda node: heapq.heappush(self.open, node)
         self.extract = lambda: heapq.heappop(self.open)
     elif search_strategy == _ASTAR:
         #use priority queue for OPEN (first out is node with
         #lowest fval = gval+hval)
         self.open = []
         #set node less than function to compare sums of hval and gval
         sNode.lt_type = _SUM_HG
         self.insert = lambda node: heapq.heappush(self.open, node)
         self.extract = lambda: heapq.heappop(self.open)
Пример #15
0
def generate_huffman_table(frequencies):

	encoding_table = {}
	decoding_table = {}

	for position_id, histogram in frequencies.iteritems():
		encoding_table[position_id] = [None] * 256
	
		# create a mapping table for characters 
		ordinal_freq = [(histogram["%02x" % ordinal], ordinal) if histogram.has_key("%02x" % ordinal) else (1, ordinal) for ordinal in xrange(256)]

		heap = []
		for node in ordinal_freq:
			heapq.heappush(heap, node)
	
		while len(heap) > 1:
			left = heapq.heappop(heap)
			right = heapq.heappop(heap)
		
			new_node = (left[0] + right[0], (left[1], right[1]))

			heapq.heappush(heap, new_node)
	
		total, huffman = heap[0]
	
		fill_table(huffman, "", encoding_table[position_id])
		decoding_table[position_id] = huffman

	return encoding_table, decoding_table
Пример #16
0
def first_k_most_relevant(doc_scores):
	"""If there are more than k documents containing terms in a query, return the k documents with the highest scores, tiebroken by least docID first.
	If there are less than k documents, return them, sorted by highest scores, and tiebroken by least docID first. O(n) + O(k lg n)

	:param doc_scores: A dictionary of docID to its corresponding document's score.
	:return: List of k docIDs of string type, which are the most relevant (i.e. have the highest scores)
	"""
	scores = [(-score, docID) for docID, score in doc_scores.iteritems()] # invert the scores so that heappop gives us the smallest score
	heapq.heapify(scores)
	most_relevant_docs = []
	for _ in range(k):
		if not scores:
			break
		most_relevant_docs.append(heapq.heappop(scores))
	if not most_relevant_docs:
		return most_relevant_docs
	# deals with equal-score cases
	kth_score, kth_docID = most_relevant_docs[-1]
	while scores:
		next_score, next_docID = heapq.heappop(scores)
		if next_score == kth_score:
			most_relevant_docs.append((next_score, next_docID))
		else:
			break
	return sort_relevant_docs(most_relevant_docs)
Пример #17
0
def streaming_median(stream):
    ''' Given an umlimited stream of numbers, compute
    the exact median of the values:

    :param stream: The stream to observe
    :returns: The median of the stream

    >>> streaming_median(range(101))
    50

    :param stream: The stream of incoming numbers
    :returns: The median of the given stream
    '''
    max_heap, min_heap = [], [] # max_heap is a min heap holding the large values
    for el in stream:           # min_heap is a max heap holding the small values
        if min_heap and el > min_heap[0]:
            heapq.heappush(min_heap, el)
        else: heapq.heappush(max_heap, -el)

        if abs(len(min_heap) - len(max_heap)) > 1:
            if len(min_heap) > len(max_heap):
                heapq.heappush(max_heap, -heapq.heappop(min_heap))
            else:
                heapq.heappush(min_heap, -heapq.heappop(max_heap))

    if   len(min_heap) > len(max_heap): return min_heap[0]
    elif len(max_heap) > len(max_heap): return -max_heap[0]
    else: return (min_heap[0] - max_heap[0]) / 2
Пример #18
0
    def dijkstra(self, node_from):
        graph = type(self)(copy.deepcopy(self.nodes), copy.deepcopy(self.vertex_map))
        # Set the distance for the start node to zero 
        graph.set_node_distance(node_from, 0)

        # Put tuple pair into the priority queue
        unvisited_queue = graph.get_nodes_unvisited()
        heapq.heapify(unvisited_queue)

        while len(unvisited_queue):
            distance, node_current = heapq.heappop(unvisited_queue)
            graph.set_node_visited(node_current)
            del distance

            for node_adj in graph.vertex_map.get(node_current, {}).keys():
                if graph.vertex_map[node_current][node_adj].get('visited', False):
                    continue
                new_dist = graph.get_node_distance(node_current) + graph.vertex_map[node_current][node_adj]['distance']
                if new_dist < graph.get_node_distance(node_adj):
                    graph.set_node_distance(node_adj, new_dist)
                    graph.set_node_previous(node_adj, node_current)

            # Rebuild heap
            # 1. Pop every item
            while len(unvisited_queue):
                heapq.heappop(unvisited_queue)

            # 2. Put all vertices not visited into the queue
            unvisited_queue = graph.get_nodes_unvisited()
            heapq.heapify(unvisited_queue)

        return graph
Пример #19
0
    def _merge_file_iters(self, iters, key=lambda x: x):
        heap = []
        for iter, root in iters:
            try:
                heap.append((next(iter), iter, root))
            except StopIteration:
                pass

        heapq.heapify(heap)
        while heap:
            value, iter, root = heapq.heappop(heap)
            try:
                heapq.heappush(heap, (next(iter), iter, root))
            except StopIteration:
                pass

            items = [(value, root)]
            while heap:
                cand, iter, root = heapq.heappop(heap)
                if key(cand) == key(value):
                    items.append((cand, root))
                    try:
                        heapq.heappush(heap, (next(iter), iter, root))
                    except StopIteration:
                        pass
                else:
                    heapq.heappush(heap, (cand, iter, root))
                    break

            yield items
Пример #20
0
 def pop(self):
     return_item = heappop(self)[1]
     while return_item not in self._item_set:
         return_item = heappop(self)[1]
     self._item_set.remove(return_item)
     self.sweep()
     return return_item
Пример #21
0
    def findMedian(self):
        """
        Returns the median of current data stream
        :rtype: float
        """
        # If odd total len, then the value on lower heap
        # If even total len, avg of both heaps.
        res = []
        if len(self.lower) > 0:
            res.append(-heapq.heappop(self.lower))
        if len(self.higher) > 0:
            res.append(heapq.heappop(self.higher))
        if len(res) > 0:
            heapq.heappush(self.lower, -res[0])
        if len(res) > 1:
            heapq.heappush(self.higher, res[1])
        if len(res) > 0:
            if len(self.lower) > len(self.higher):
                return float(res[0])
            else:
                return (res[0] + res[1])/float(2)
        return -0.0

# Your MedianFinder object will be instantiated and called as such:
# mf = MedianFinder()
# mf.addNum(1)
# mf.findMedian()
Пример #22
0
    def minMeetingRooms(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: int
        """
        """
        用一个min-heap, 非空且下一个interval.start >= 上一个的end时, 意味不需要another room,
        pop; 不然, 需要一个新的room.  heap里按end比较,所以存的是(interval.end, interval)
        """
        import heapq

        if not intervals:
            return 0

        intervals = sorted(intervals, key=lambda x: x.start)

        queue = []
        max_len = 1
        for i in xrange(len(intervals)):
            while queue and intervals[i].start >= queue[0][0]:
                heapq.heappop(queue)
            heapq.heappush(queue, (intervals[i].end, intervals[i]))  # 注意heap中用end来比较
            max_len = max(max_len, len(queue))

        return max_len
Пример #23
0
def print_topK():
	print "size of priority q" + str(len(topKQ))
	for item in topKQ:
		(val,key) = heapq.heappop(topKQ)
		print key, "st \t", int(val)	
	(val,key) = heapq.heappop(topKQ)
	print key, "st \t", int(val)	
Пример #24
0
    def run(self):
        headers = bam_headers(self.input)
        writer = Bam_writer(self.prefix+'.bam', headers)
        chrom = None
        endpoints = [ ]
        total = 0
        discarded = 0
        for record in Bam_reader(self.input):
            if record.flag & FLAG_UNMAPPED: continue
        
            total += 1
            
            if chrom != record.rname:
                chrom = record.rname
                endpoints = [ ]
            while endpoints and endpoints[0] <= record.pos:
                heapq.heappop(endpoints)
            
            if len(endpoints) >= self.depth:
                discarded += 1
                continue

            heapq.heappush(endpoints, record.pos+record.length)
            
            record.flag &= ~(FLAG_PAIRED|FLAG_PROPER|FLAG_MATE_UNMAPPED|FLAG_MATE_REVERSE|FLAG_FIRST|FLAG_SECOND)
            record.mrnm = '*'
            record.mpos = 0
            writer.write(record)
        writer.close()
        
        self.log.log('Discarded %s alignments out of %s.\n' % (grace.pretty_number(discarded),grace.pretty_number(total)))
Пример #25
0
    def _query(self, p, k=1, eps=0, distance_upper_bound=np.inf):
        if not self.root:
            return []

        dist_to_ctr = self.distance(p, self.data[self.root.ctr_idx])
        min_distance = max(0.0, dist_to_ctr - self.root.radius)

        # priority queue for chasing nodes
        # entries are:
        #  minimum distance between the node area and the target
        #  distance between node center and target
        #  the node
        q = [(min_distance,
              dist_to_ctr,
              self.root)]
        # priority queue for the nearest neighbors
        # furthest known neighbor first
        # entries are (-distance, i)
        neighbors = []

        if eps == 0:
            epsfac = 1
        else:
            epsfac = 1 / (1 + eps)

        while q:
            min_distance, dist_to_ctr, node = heappop(q)
            if isinstance(node, CoverTree._LeafNode):
                # brute-force
                for i in node.idx:
                    if i == node.ctr_idx:
                        d = dist_to_ctr
                    else:
                        d = self.distance(p, self.data[i])
                    if d <= distance_upper_bound:
                        if len(neighbors) == k:
                            heappop(neighbors)
                        heappush(neighbors, (-d, i))
                        if len(neighbors) == k:
                            distance_upper_bound = -neighbors[0][0]
            else:
                # we don't push nodes that are too far onto the queue at
                # all, but since the distance_upper_bound decreases, we
                # might get here even if the cell's too far
                if min_distance > distance_upper_bound * epsfac:
                    # since this is the nearest node, we're done, bail out
                    break

                for child in node.children:
                    if child.ctr_idx == node.ctr_idx:
                        d = dist_to_ctr
                    else:
                        d = self.distance(p, self.data[child.ctr_idx])
                    min_distance = max(0.0, d - child.radius)

                    # child might be too far, if so, don't bother pushing it
                    if min_distance <= distance_upper_bound * epsfac:
                        heappush(q, (min_distance, d, child))

        return sorted([(-d, i) for (d, i) in neighbors])
Пример #26
0
    def _merge_terms(self, iterlist):
        # Merge-sorts terms coming from a list of term iterators.

        # Create a map so we can look up each iterator by its id() value
        itermap = {}
        for it in iterlist:
            itermap[id(it)] = it

        # Fill in the list with the head term from each iterator.

        current = []
        for it in iterlist:
            term = next(it)
            current.append((term, id(it)))
        heapify(current)

        # Number of active iterators
        active = len(current)
        while active:
            # Peek at the first term in the sorted list
            term = current[0][0]

            # Re-iterate on all items in the list that have that term
            while active and current[0][0] == term:
                it = itermap[current[0][1]]
                try:
                    nextterm = next(it)
                    heapreplace(current, (nextterm, id(it)))
                except StopIteration:
                    heappop(current)
                    active -= 1

            # Yield the term
            yield term
    def isPossible(self, nums: List[int]) -> bool:
        """
        (length, last)
        heap sortest first
        >= 3, then drop

        split when duplicate
        """
        h = []
        for n in nums:
            while h and h[0].end + 1 < n:
                itvl = heapq.heappop(h)
                if itvl.length < 3:
                    return False

            if not h:
                heapq.heappush(h, Interval(n, 1))
            elif h[0].end + 1 == n:
                itvl = heapq.heappop(h)
                heapq.heappush(h, Interval(n, itvl.length + 1))
            else:  # n == end
                heapq.heappush(h, Interval(n, 1))


        for itvl in h:
            if itvl.length < 3:
                return False

        return True
Пример #28
0
 def pump(self, *arg, **kwarg):
     """
     Pumps all events that are applicable to be called.
     """
     # TODO: Reconfigure to use a number provided in pump method arguments
     while self.event_queue and self.event_queue[0][0] <= video.get_time():
         heapq.heappop(self.event_queue)[1](*arg, **kwarg)
Пример #29
0
def heap_merged(items_lists, combiner):
    heap = []
    def pushback(it):
        try:
            k,v = it.next()
            # put i before value, so do not compare the value
            heapq.heappush(heap, (k, i, v))
        except StopIteration:
            pass
    for i, it in enumerate(items_lists):
        if isinstance(it, list):
            items_lists[i] = it = (k for k in it)
        pushback(it)
    if not heap: return

    last_key, i, last_value = heapq.heappop(heap)
    pushback(items_lists[i])

    while heap:
        k, i, v = heapq.heappop(heap)
        if k != last_key:
            yield last_key, last_value
            last_key, last_value = k, v
        else:
            last_value = combiner(last_value, v)
        pushback(items_lists[i])

    yield last_key, last_value
Пример #30
0
 def reorganizeString(self, S):
     """
     :type S: str
     :rtype: str
     """
     if len(S) <= 1:
         return S
     l, c = len(S), [[-item[1], item[0]] for item in Counter(S).items()] + [[1, '#']]
     heapq.heapify(c)
     rls = ''
     for i in range(l):
         t1 = heapq.heappop(c)
         t2 = heapq.heappop(c)
         if rls == '' or t1[1] != rls[-1]:
             rls += t1[1]
             t1[0] += 1
         else:
             if t2[1] == '#':
                 return ''
             rls += t2[1]
             t2[0] += 1
         if t1[0] != 0:
             heapq.heappush(c, t1)
         if t2[0] != 0:
             heapq.heappush(c, t2)
     return rls
Пример #31
0
# https://www.acmicpc.net/problem/2211
# 이름 : 네트워크 복구
# 번호 : 2211
# 난이도 : 골드 II
# 분류 : 그래프 이론, 다익스트라

import sys
from heapq import heappush, heappop
input = sys.stdin.readline
INF = sys.maxsize
N, M = map(int, input().split())
graph = [[] for _ in range(N + 1)]
for _ in range(M):
    A, B, C = map(int, input().split())
    graph[A].append((C, B))
    graph[B].append((C, A))
ans = [''] * (N + 1)
dp = [INF] * (N + 1)
dp[1] = 0
heap = []
heappush(heap, [0, 1])
while heap:
    dist, nx = heappop(heap)
    for n_dist, x in graph[nx]:
        temp = dist + n_dist
        if dp[x] > temp:
            dp[x] = temp
            ans[x] = str(nx) + ' ' + str(x)
            heappush(heap, [temp, x])
print(N - 1)
print('\n'.join(ans[2:]))
Пример #32
0
n = int(input())
a = [-int(j) for j in input().split()]
a.sort()
import heapq
l = [a[0]]
heapq.heapify(l)
ans = 0
for i in a[1:]:
    heapq.heappush(l, i)
    heapq.heappush(l, i)
    p = heapq.heappop(l)
    ans -= p
print(ans)
Пример #33
0
queue[0] # return 2 examine the first element

## Stack
stack = []
size = len(stack)
stack.append(1)
stack.append(2)
stack.pop() # return 2

## Priority Queue - https://docs.python.org/2/library/heapq.html
import heapq
heap = []
heapq.heappush(heap, 1)  # heap [1]
heapq.heappush(heap, 3)  # heap [1,3]
heapq.heappush(heap, 2)  # heap [1,3,2]
heapq.heappop(heap)  # 1
heapq.heappop(heap)  # 2
heapq.heappop(heap)  # 3

## Deque
import collections
dq = collections.deque()
dq.append(1)  # [1]
dq.appendleft(2)  # [2,1]
dq.popleft()  # [1]


## Set
s = set({1,3,4,5,6,5})  # {1, 3, 4, 5, 6}

## List
Пример #34
0
def encode(file):
    """Encodes file into a smaller representation using huffman encoding.
    The encoded file is written to the filename "compressed.txt"
    Params: file - str
    """
    # open the file
    try:
        s = open(file).read()
    except IOError:
        print("Error opening file: {}".format(file))
        exit(1)

    # build dict of character: frequency
    freqs = make_dict(s)

    # build the Huffman Tree.
    # 1. Make leaf nodes, and store the letter and frequency.
    # 2. Build a tree by taking the 2 nodes with the smallest frequency, and creating a parent with the
    #    letters concatenated and the frequencies summed.
    # I used a minheap for log(N) access to the 2 nodes with the smallest frequency each time I pop from the heap.
    minheap = []
    for k, v in freqs.items():
        heappush(minheap, TreeNode(k, v))

    # ASSUMPTION: since the bitmap is of size 100 x 100 it must be separated by newlines \n,
    # so there are at least 2 distinct characters in any input file.
    parent_node = None # save the current parent node
    # This is step 2 - building the tree by creating parent nodes and keeping track of child pointers.
    while True:
        node1, node2 = heappop(minheap), heappop(minheap)
        parent_node = TreeNode(node1.letter + node2.letter, node1.freq + node2.freq)
        parent_node.left = node1
        parent_node.right = node2
        if parent_node.freq == len(s):
            break
        else:
            heappush(minheap, parent_node)

    # We know the tree is built when a parent node's frequency equals the length of the original string.
    # 3. Recursively iterate through the tree, creating encodings for all of the leaf nodes.
    # Since there is a unique path from the root to any particular node, each encoding will be unique.
    leaves = []
    tree_dfs(parent_node, "", leaves)
    encoding_to_char = {}
    for leaf in leaves:
        assert len(leaf.letter) == 1
        assert leaf.encoding not in encoding_to_char # because the encodings should have been unique.
        encoding_to_char[leaf.encoding] = leaf.letter

    # 4. Create the dictionary that maps characters in our original file to their encodings.
    char_to_encoding = {v: k for k, v in encoding_to_char.items()}

    # Representation of the encoded file. First we convert it into a binary string and then hex to save space.
    int_str = "".join([char_to_encoding[char] for char in s])
    hex_str = hex(int(int_str, 2))

    # hack to fix in the future - This way of getting the hex form does not preserve leading zeros,
    # leading to occasional inaccuracy in converting back and forth between binary and hex.
    num_zeros_to_add = 0
    bin_str = bin(int(hex_str, 16))[2:]
    while bin_str != int_str:
        bin_str  = '0' + bin_str
        num_zeros_to_add+=1

    # 5. Write out the compressed file. We will need both the string and dictionary to dencode at a later time.

    try:
        os.remove('compressed.txt')
    except OSError:
        pass

    with open('compressed.txt', 'wb') as file:
        pickle.dump((hex_str, char_to_encoding, num_zeros_to_add), file)

    # A sanity check to make sure that the mapping and strings were written correctly.
    with open('compressed.txt', 'rb') as file:
        hs, mapping, add_zero = pickle.load(file)

    assert mapping == char_to_encoding, "houston we have a problem"
    assert bin_str == int_str, "{} {}".format(bin_str, int_str)
Пример #35
0
    def run(self, run_to, node_state_grid=None,
            plot_each_transition=False, plotter=None):
        """Run the model forward for a specified period of time.

        Parameters
        ----------
        run_to : float
            Time to run to, starting from self.current_time
        node_state_grid : 1D array of ints (x number of nodes) (optional)
            Node states (if given, replaces model's current node state grid)
        plot_each_transition : bool (optional)
            Option to display the grid after each transition
        plotter : CAPlotter object (optional)
            Needed if caller wants to plot after every transition
        """
        if node_state_grid is not None:
            self.set_node_state_grid(node_state_grid)
       
        # Continue until we've run out of either time or events
        while self.current_time < run_to and self.event_queue:

            if _DEBUG:
                print('Current Time = ', self.current_time)

            # Is there an event scheduled to occur within this run?
            if self.event_queue[0].time <= run_to:

                # If so, pick the next transition event from the event queue
                ev = heappop(self.event_queue)

                if _DEBUG:
                    print('Event:', ev.time, ev.link, ev.xn_to)
    
                # ... and execute the transition
                if _USE_CYTHON:
                    do_transition(ev, self.next_update,
                                  self.grid.node_at_link_tail,
                                  self.grid.node_at_link_head,
                                  self.node_state, self.link_state,
                                  self.san, self.link_orientation,
                                  self.propid, self.prop_data,
                                  self.n_xn, self.xn_to, self.xn_rate,
                                  self.grid.links_at_node,
                                  self.grid.active_link_dirs_at_node,
                                  self.num_node_states, self.num_node_states_sq,
                                  self.prop_reset_value, self.xn_propswap,
                                  self.xn_prop_update_fn,
                                  self.bnd_lnk, self.event_queue,
                                  self,
                                  plot_each_transition,
                                  plotter)
                else:
                    self.do_transition(ev, self.current_time,
                                       plot_each_transition, plotter)

                # Update current time
                self.current_time = ev.time
                
            # If there is no event scheduled for this span of time, simply
            # advance current_time to the end of the current run period.
            else:
                self.current_time = run_to
Пример #36
0
 def pop(self):
     (priority, _, item) = heapq.heappop(self.heap)
     return (item, priority)
Пример #37
0
import heapq
import sys
input = sys.stdin.readline

T, M, K = map(int, input().split())
KKK = list(map(int, input().split()))
arr = [list(map(int, input().split())) for i in range(M)]
dic = defaultdict(list)
for i, j, cost in arr:
    dic[i].append((cost, j))
    dic[j].append((cost, i))
Q = []
for n in KKK:
    Q.extend(dic[n][:])
heapq.heapify(Q)
visit = [False] * (T + 1)
for n in KKK:
    visit[n] = True
res = 0
size = K
while Q:
    if size == T:
        break
    cost, next = heapq.heappop(Q)
    if visit[next] == False:
        visit[next] = True
        res += cost
        for i in dic[next]:
            heapq.heappush(Q, i)
        size += 1
print(res)
Пример #38
0
import heapq
import sys

N = int(input())
colors = list(map(int, input().strip().split()))
heapq.heapify(colors)

if (len(colors) % 2 == 0):
    colors.append(0)

ans = 0

while (len(colors) > 2):
    a = heapq.heappop(colors)
    b = heapq.heappop(colors)
    c = heapq.heappop(colors)
    heapq.heappush(colors, a + b + c)
    ans += a + b + c

print(ans)
Пример #39
0
def crawl_next(inlink, frontier, time_record, url_crawled, outlink,
               url_response):
    next_frontier = set()
    frontier_size = len(frontier)
    url_crawled_this_wave = set()
    while frontier:
        try:
            (priority, url, parent) = heapq.heappop(frontier)
        except Exception as ex:
            continue

        # 1. Check if this is a good url, if not, skip
        if url in url_crawled:
            if url in url_crawled_this_wave:
                inlink[url].append(parent)
            continue

        if check_media(url):
            continue
        try:
            # response = request.urlopen(url, timeout=5)
            restriction = {"Accept-Language": "en-US,en;q=0.5"}
            response = requests.get(url, headers=restriction, timeout=5)
            url_component = parse.urlparse(url)
        except Exception as ex:
            continue
        domain = url_component.netloc.lower()
        scheme = url_component.scheme.lower()
        if domain.split(
                '.')[1] == 'wikipedia' and domain.split('.')[0] != 'en':
            continue
        if len(domain.split('.')) > 2:
            if domain.split('.')[2] == 'gov':
                continue
        if not check_robot(scheme, domain, url):
            continue
        if not check_response(response):
            continue
        print("Crawled: " + str(len(url_crawled)) + " URL. Currently on: " +
              url)

        # 2. Get the next urls
        # next_urls = get_next_urls(url, response)
        if domain not in time_record:
            time_record[domain] = time.time()
            next_urls = get_next_urls(url, response)
        else:
            if (time.time() - time_record[domain]) < 1:
                print("Sleep for" + str((time.time() - time_record[domain])))
                time.sleep(time.time() - time_record[domain])
                next_urls = get_next_urls(url, response)
            else:
                time_record[domain] = time.time()
                next_urls = get_next_urls(url, response)

        # 3. Saving information of the urls
        url_response[url] = response
        url_crawled.add(url)
        if parent != None:
            inlink[url].append(parent)
        else:
            if url in inlink:
                pass
            else:
                inlink[url] = []
        for next_url in next_urls:
            next_frontier.add((next_url, url))
            if next_url in url_crawled:
                if str(url) != str(next_url):
                    inlink[next_url].append(url)
        next_urls = list(next_urls)
        outlink[url] = next_urls

        # 4. Dump the web content when necessary.
        if (len(url_crawled) % 10) == 0:
            dump_inlink(inlink)
            dump_url_crawled(url_crawled)
            dump_frontier(frontier)
            dump_content(outlink, url_response)
        # Only process top 60% scored urls.
        if (len(frontier) / frontier_size) < 0.4 and len(frontier) > 6:
            break

        # 5. Stop and exit when we crawled enough urls.
        if len(url_crawled) >= 40300:
            dump_inlink(inlink)
            print("Finish crawling")
            exit()
    return next_frontier
import sys, heapq
sys.stdin = open('백준1937. 욕심쟁이 판다.txt', 'r')

d = [(0, 1), (0, -1), (1, 0), (-1, 0)]
n = int(input())
ground = [[0] * n for _ in range(n)]
matrix = [list(map(int, sys.stdin.readline().split())) for _ in range(n)]
heap = []

for r in range(n):
    for c in range(n):
        heapq.heappush(heap, (matrix[r][c], r, c))

answer = 0
while heap:
    bamboo, r, c = heapq.heappop(heap)
    for idx in range(4):
        nr = r + d[idx][0]
        nc = c + d[idx][1]
        if 0 <= nr < n and 0 <= nc < n:
            if matrix[r][c] > matrix[nr][nc]:
                ground[r][c] = max(ground[r][c], ground[nr][nc])
    ground[r][c] += 1
    answer = max(answer, ground[r][c])

print(answer)

# DFS
# import sys
# input = sys.stdin.readline
# sys.setrecursionlimit(10**6)
Пример #41
0
def find_path(vertex1, vertex2, external_path):
    # pdb.set_trace()
    priority_queue = []
    visited = set()
    v1 = PriorityQueueVertex(vertex1, 0, None)

    v1_key = Graph.compute_heuristic(vertex1, vertex2)
    heapq.heappush(priority_queue, (v1_key, v1))
    visited.add(vertex1)

    # start search
    path = []
    destination_vertex = vertex2

    occupied_edges = set()
    while priority_queue[0][1] != destination_vertex:
        # pdb.set_trace()
        current_v = priority_queue[0][1]
        current_cost = current_v.cost

        visited.add(current_v.vertex)

        # update edges that are currently occupied
        for e in occupied_edges:
            e.cost = Edge.UNOCCUPIED_COST

        # set occupied edges
        occupied_edges = set()
        for i in range(1, 3):
            try:
                occupied_vertex = external_path[current_v.depth + i]
            except IndexError:
                continue
            if i == 1:
                for e in occupied_vertex.edges:
                    if e is not None:
                        other_edge = occupied_vertex.get_opposite_edge(e)
                        if other_edge is not None:
                            occupied_edges.add(other_edge)
            if i == 2:
                # update wait edge
                e = occupied_vertex.edges[4]
                occupied_edges.add(e)

        for e in occupied_edges:
            e.cost = Edge.OCCUPIED_COST

        wait_flag = False
        for e in current_v.vertex.edges:
            if e is not None and e.cost == Edge.OCCUPIED_COST:
                wait_flag = True
                break

        for e in current_v.vertex.edges:
            if e is None:
                continue
            # calculate cost and heuristics, push onto PQ
            v = e.destination

            if wait_flag:
                if v in visited and v != current_v.vertex:
                    continue
            else:
                if v in visited:
                    continue

            v_cost = current_cost + e.cost
            v_key = (v_cost + Graph.compute_heuristic(v, destination_vertex))
            v_priority = PriorityQueueVertex(v, v_cost, current_v)
            heapq.heappush(priority_queue, (v_key, v_priority))

        if wait_flag:
            # there might be duplicate entries in the priority queue.
            # Only remove the smallest one
            holder = [i for i in priority_queue if i[1] == current_v]
            if len(holder) > 1:
                holder.sort()
                v_key = holder[0][0]
                priority_queue.remove((v_key, current_v))
        else:
            priority_queue = [i for i in priority_queue if i[1] != current_v]
        heapq.heapify(priority_queue)

    # pdb.set_trace()
    # Now get path
    v_top = heapq.heappop(priority_queue)
    v = v_top[1]
    path = []
    while True:
        path.append(v.vertex)
        v = v.previous_vertex
        if v is None:
            break
    path.reverse()

    return path
Пример #42
0
import heapq

for sadsf in range(int(input())):
    n, m = map(int, input().split())
    m -= 1
    h = []
    heapq.heappush(h, -n)
    d = {n: 1}
    j = n
    while m > 0:
        #        print(d, h)
        if d[j] > m:
            break
        m -= d[j]
        if (j - 1) // 2 not in d:
            d[(j - 1) // 2] = 0
            heapq.heappush(h, -((j - 1) // 2))
        d[(j - 1) // 2] += d[j]
        if j // 2 not in d:
            d[j // 2] = 0
            heapq.heappush(h, -(j // 2))
        d[j // 2] += d[j]
        del d[j]
        heapq.heappop(h)
        j = -h[0]

    print('Case #{}: {} {}'.format(sadsf + 1, j // 2, (j - 1) // 2))
Пример #43
0
def main(_):
    #initial outer file
    WordIndex = Word_Index()
    EntTagIndex = Ent_Tag_Index()
    TriTagIndex = Tri_Tag_Index()
    PtagIndex = Ptag_Index()
    Index = [EntTagIndex, TriTagIndex]
    object_tag_Index = Index[FLAGS.object_id]

    FLAGS.num_word = len(WordIndex.word2idex)
    FLAGS.num_postag = len(PtagIndex.ptag2id)
    FLAGS.num_class = object_tag_Index.num_class
    df_test = pd.read_csv('datas/corpus/testdata_id.txt',
                          sep='#',
                          skip_blank_lines=False,
                          dtype={'len': np.int32})
    df_train = pd.read_csv('datas/corpus/traindata_id.txt',
                           sep='#',
                           skip_blank_lines=False,
                           dtype={'len': np.int32})
    # eval_size = int(len(df_train) * FLAGS.dev_rate)
    # df_eval = df_train.iloc[-eval_size:]
    # df_train = df_train.iloc[:-eval_size]
    print('trainsize' + str(len(df_train)))
    train_data_itor = DataItor(df_train, False)
    # eval_data_itor = DataItor(df_eval,False)
    test_data_itor = DataItor(df_test, False)
    FLAGS.check_every_point = int(train_data_itor.size / FLAGS.batch_size)
    if os.path.exists(FLAGS.pretrain_file) and FLAGS.use_pretrain_embedding:
        print('initial!!!!!!!!!!')
        # initial_embedding(WordIndex.word2idex)
        FLAGS.pretrain_emb = initial_embedding(WordIndex.word2idex)
    else:
        FLAGS.pretrain_emb = None

    myconfig = tf.ConfigProto(allow_soft_placement=True)
    with tf.Session(config=myconfig) as sess:
        model = ST_Model(FLAGS)
        sess.run(tf.global_variables_initializer())
        # xidx_eval, tidx_eval, lens_eval = eval_data_itor.next_all()
        xidx_test, tidx_test, lens_test = test_data_itor.next_all()
        # print('eval data size %f' % len(tidx_eval[0]))

        # _, id2label = helper.loadMap(FLAGS.label2id_path)
        saver = tf.train.Saver(max_to_keep=2)
        previous_best_valid_f1_score = 0
        previous_best_epoch = -1
        bad_count = 0
        heap_source, heap_trigger, heap_target = [], [], []
        while train_data_itor.epoch < FLAGS.num_epochs:
            x_train_batch, y_train_batch = train_data_itor.next_batch(
                FLAGS.batch_size)

            train_step, train_loss = model.train_model(
                sess, x_train_batch, y_train_batch[FLAGS.object_id])

            if train_data_itor.batch_time % FLAGS.check_every_point == 0:
                print("current batch_time: %d" % (train_data_itor.batch_time))
                # y_eval_pred, eval_loss = model.inference_for_single(sess, xidx_eval, tidx_eval[FLAGS.object_id])
                y_test_pred, test_loss = model.inference_for_single(
                    sess, xidx_test, tidx_test[FLAGS.object_id])
                if FLAGS.object_id == 0:
                    # precison, recall, f1_eval = helper.evaluate_ent(xidx_eval[0], tidx_eval[FLAGS.object_id],
                    #                                                y_eval_pred, id2word=WordIndex.id2word,id2label=EntTagIndex.id2tag,seq_lens=lens_eval)
                    # print('evalution on eval data, target_eval_loss:%.4f,precison:%.4f,recall:%.4f,fscore:%.4f' % (
                    # eval_loss, precison, recall, f1_eval))
                    precison1, recall1, f1_test = helper.evaluate_ent(
                        xidx_test[0],
                        tidx_test[FLAGS.object_id],
                        y_test_pred,
                        id2word=WordIndex.id2word,
                        id2label=EntTagIndex.id2tag,
                        seq_lens=lens_test)
                    print(
                        'evalution on test data, target_eval_loss:%.3f,precison:%.4f,recall:%.4f,fscore:%.4f'
                        % (test_loss, precison1, recall1, f1_test))
                else:
                    # precison, recall, f1_eval =  helper.evaluate_tri(xidx_eval[0], tidx_eval[FLAGS.object_id], y_eval_pred,id2word=WordIndex.id2word,seq_lens=lens_eval)
                    # print('evalution on eval data, target_eval_loss:%.3f,precison:%.4f,recall:%.4f,fscore:%.4f' % (
                    # eval_loss, precison, recall, f1_eval))
                    precison1, recall1, f1_test = helper.evaluate_tri(
                        xidx_test[0],
                        tidx_test[FLAGS.object_id],
                        y_test_pred,
                        id2word=WordIndex.id2word,
                        seq_lens=lens_test)
                    print(
                        'evalution on test data, target_eval_loss:%.3f,precison:%.4f,recall:%.4f,fscore:%.4f'
                        % (test_loss, precison1, recall1, f1_test))

                # early stop
                if len(heap_target) < 5:
                    heapq.heappush(heap_target,
                                   (f1_test, (precison1, recall1, f1_test)))
                else:
                    if f1_test > heap_target[0][0]:
                        heapq.heappop(heap_target)
                        heapq.heappush(heap_target,
                                       (f1_test,
                                        (precison1, recall1, f1_test)))
                if f1_test > previous_best_valid_f1_score:
                    helper.store_stl_result(
                        xidx_test[0],
                        tidx_test[FLAGS.object_id],
                        y_test_pred,
                        id2word=WordIndex.id2word,
                        id2label=Index[FLAGS.object_id].id2tag,
                        seq_lens=lens_test,
                        type_id=FLAGS.object_id)

                    previous_best_valid_f1_score = f1_test
                    bad_count = 0
                else:
                    bad_count += 1

                if bad_count >= FLAGS.patients:
                    print('early stop!')
                    break
        print('Train Finished!!')
        show_result(heap_target)

    pass
Пример #44
0
 def get(self):
   return heapq.heappop(self.elements)[1]
Пример #45
0
graph = defaultdict(list)
for u, v, w in data:
    graph[u].append([v, w])

# perform
INF = 3000000
result = [INF for _ in range(V)]

heap = []
heapq.heappush(heap, (0, start_point))

result[start_point-1] = 0

while heap:
    cur_weight, cur_point = heapq.heappop(heap)

    if cur_weight > result[cur_point-1]:
        print(cur_weight, result[cur_point-1])
        continue

    for pos, wei in graph[cur_point]:
        tmp = cur_weight + wei
        if tmp < result[pos-1]:
            result[pos-1] = tmp
            heapq.heappush(heap, (tmp, pos))

# output
for r in result:
    if r == INF:
        print("INF")
Пример #46
0
 def get(self): # Pop and return the smallest item from the heap
     return heapq.heappop(self.elements)[1]
Пример #47
0
 def pop(self):
     return heappop(self.q)
Пример #48
0
#21.02.02
import sys
import heapq
input = sys.stdin.readline

N = int(input())
rooms,query = [],[]
for _ in range(N):
    query.append(list(map(int,input().strip().split())))
query=sorted(query,key=lambda x:x[0])
for i in range(N):
    # print("i :",i,"rooms :",rooms,"Query :",query)
    if len(rooms)!=0 and rooms[0]<=query[i][0]:
        heapq.heappop(rooms)
    heapq.heappush(rooms,query[i][1])
print(len(rooms))
Пример #49
0
import heapq
n, k1, k2 = map(int, input().split())
a = list(map(int, input().split()))[:n]
b = list(map(int, input().split()))[:n]
total = k1 + k2
arr = []
for i in range(n):
    arr.append((-1) * abs(a[i] - b[i]))
heapq.heapify(arr)
while (total > 0):
    item = (-1) * (heapq.heappop(arr))
    heapq.heappush(arr, (-1) * abs(item - 1))
    total -= 1
ans = 0
for i in range(len(arr)):
    ans += (arr[i])**2
print(ans)
Пример #50
0
 def pop_smallest(self):
     """ Remove and return the smallest item from the queue
     """
     smallest = heapq.heappop(self.heap)
     del self.set[smallest]
     return smallest
Пример #51
0
import heapq

# heapq.heappush()
# heapq.heappop()
# list를 heap구조로 사용하는 함수.

N = int(input())

heap = list()
result = list()

for _ in range(N):
    data = int(input())
    if data == 0:
        if heap:
            result.append(heapq.heappop(heap))
        else:
            result.append(0)
    else:
        heapq.heappush(heap, data)

for data in result:
    print(data)
Пример #52
0
    def start(self):
        if not logging.getLogger().handlers:
            # The IOLoop catches and logs exceptions, so it's
            # important that log output be visible.  However, python's
            # default behavior for non-root loggers (prior to python
            # 3.2) is to print an unhelpful "no handlers could be
            # found" message rather than the actual log entry, so we
            # must explicitly configure logging if we've made it this
            # far without anything.
            logging.basicConfig()
        if self._stopped:
            self._stopped = False
            return
        old_current = getattr(IOLoop._current, "instance", None)
        IOLoop._current.instance = self
        self._thread_ident = thread.get_ident()
        self._running = True

        # signal.set_wakeup_fd closes a race condition in event loops:
        # a signal may arrive at the beginning of select/poll/etc
        # before it goes into its interruptible sleep, so the signal
        # will be consumed without waking the select.  The solution is
        # for the (C, synchronous) signal handler to write to a pipe,
        # which will then be seen by select.
        #
        # In python's signal handling semantics, this only matters on the
        # main thread (fortunately, set_wakeup_fd only works on the main
        # thread and will raise a ValueError otherwise).
        #
        # If someone has already set a wakeup fd, we don't want to
        # disturb it.  This is an issue for twisted, which does its
        # SIGCHILD processing in response to its own wakeup fd being
        # written to.  As long as the wakeup fd is registered on the IOLoop,
        # the loop will still wake up and everything should work.
        old_wakeup_fd = None
        if hasattr(signal, 'set_wakeup_fd') and os.name == 'posix':
            # requires python 2.6+, unix.  set_wakeup_fd exists but crashes
            # the python process on windows.
            try:
                old_wakeup_fd = signal.set_wakeup_fd(
                    self._waker.write_fileno())
                if old_wakeup_fd != -1:
                    # Already set, restore previous value.  This is a little racy,
                    # but there's no clean get_wakeup_fd and in real use the
                    # IOLoop is just started once at the beginning.
                    signal.set_wakeup_fd(old_wakeup_fd)
                    old_wakeup_fd = None
            except ValueError:  # non-main thread
                pass

        while True:
            poll_timeout = _POLL_TIMEOUT

            # Prevent IO event starvation by delaying new callbacks
            # to the next iteration of the event loop.
            with self._callback_lock:
                callbacks = self._callbacks
                self._callbacks = []
            for callback in callbacks:
                self._run_callback(callback)
            # Closures may be holding on to a lot of memory, so allow
            # them to be freed before we go into our poll wait.
            callbacks = callback = None

            if self._timeouts:
                now = self.time()
                while self._timeouts:
                    if self._timeouts[0].callback is None:
                        # the timeout was cancelled
                        heapq.heappop(self._timeouts)
                        self._cancellations -= 1
                    elif self._timeouts[0].deadline <= now:
                        timeout = heapq.heappop(self._timeouts)
                        self._run_callback(timeout.callback)
                        del timeout
                    else:
                        seconds = self._timeouts[0].deadline - now
                        poll_timeout = min(seconds, poll_timeout)
                        break
                if (self._cancellations > 512 and self._cancellations >
                    (len(self._timeouts) >> 1)):
                    # Clean up the timeout queue when it gets large and it's
                    # more than half cancellations.
                    self._cancellations = 0
                    self._timeouts = [
                        x for x in self._timeouts if x.callback is not None
                    ]
                    heapq.heapify(self._timeouts)

            if self._callbacks:
                # If any callbacks or timeouts called add_callback,
                # we don't want to wait in poll() before we run them.
                poll_timeout = 0.0

            if not self._running:
                break

            if self._blocking_signal_threshold is not None:
                # clear alarm so it doesn't fire while poll is waiting for
                # events.
                signal.setitimer(signal.ITIMER_REAL, 0, 0)

            try:
                event_pairs = self._impl.poll(poll_timeout)
            except Exception as e:
                # Depending on python version and IOLoop implementation,
                # different exception types may be thrown and there are
                # two ways EINTR might be signaled:
                # * e.errno == errno.EINTR
                # * e.args is like (errno.EINTR, 'Interrupted system call')
                if (getattr(e, 'errno', None) == errno.EINTR or
                    (isinstance(getattr(e, 'args', None), tuple)
                     and len(e.args) == 2 and e.args[0] == errno.EINTR)):
                    continue
                else:
                    raise

            if self._blocking_signal_threshold is not None:
                signal.setitimer(signal.ITIMER_REAL,
                                 self._blocking_signal_threshold, 0)

            # Pop one fd at a time from the set of pending fds and run
            # its handler. Since that handler may perform actions on
            # other file descriptors, there may be reentrant calls to
            # this IOLoop that update self._events
            self._events.update(event_pairs)
            while self._events:
                fd, events = self._events.popitem()
                try:
                    if self._handlers.has_key(fd):
                        self._handlers[fd](fd, events)
                except (OSError, IOError) as e:
                    if e.args[0] == errno.EPIPE:
                        # Happens when the client closes the connection
                        pass
                    else:
                        self.handle_callback_exception(self._handlers.get(fd))
                except Exception:
                    self.handle_callback_exception(self._handlers.get(fd))
        # reset the stopped flag so another start/stop pair can be issued
        self._stopped = False
        if self._blocking_signal_threshold is not None:
            signal.setitimer(signal.ITIMER_REAL, 0, 0)
        IOLoop._current.instance = old_current
        if old_wakeup_fd is not None:
            signal.set_wakeup_fd(old_wakeup_fd)
Пример #53
0
def irv_margin(election, winner=None, elim_order=None, ub=None, trace=False, timeout=1e75):
    '''Compute the exact IRV margin of the election.

    @type  election: L{Election}
    @param election: The election.
    @type  winner: number
    @param winner: The winner of the election. If C{None}, it will be computed.
    @type  elim_order: list
    @param elim_order: The elimination order. If C{None}, it will be computed.
    @type  ub: number
    @param ub: An upper bound on the margin. If C{None}, it will be computed via L{irv_ub}.
    @type  trace: boolean
    @param trace: If C{True}, print a trace of the computation.
    @type  timeout: number
    @param timeout: The amount of time to spend computing the margin.
    @rtype: number
    @return: Returns the IRV margin.
    '''
    if ilp is None:
        raise Exception('Cannot compute irv_margin() because no optimizer library could be loaded.')
    then = time.time()
    if winner is None or elim_order is None:
        winner, _, elim_order = irv(election)
    if ub is None:
        ub = irv_ub(election, winner=winner, elim_order=elim_order)

    # We can eliminate candidates who _must_ be eliminated first.  The
    # reasoning is that for them to not be eliminated, more votes would have to
    # be shifted than the upper bound on the margin.
    root = election.profile.deepcopy()
    while True:
        esets = []
        _elimination_set(root, all_sets=esets, rules=SF_RCV_RULES)
        max_eset = set()
        candidates = root.children()
        for eset in esets:
            if len(eset) <= len(max_eset):
                continue
            lb = min(root.get_child(c).value for c in candidates - eset) - \
                 sum(root.get_child(c).value for c in eset)
            if lb > ub:
                max_eset = eset
        if not max_eset:
            break
        for c in max_eset:
            if trace:
                print 'Eliminating candidate %d' % c
            root.eliminate(c)

    # This is the algorithm from Magrino et al., more or less
    candidates = root.children()
    k = len(candidates)
    tertiary = []
    elims = set((winner,))
    j = len(elim_order)-1
    for i in range(1, k+1):
        if i >= len(elims):
            elims = elims.union(elim_order[j])
            j -= 1
        tertiary.append(elims)
    if trace:
        print tertiary
        
    ranks = election.ranks
    fringe = []
    for c in candidates:
        if c == winner:
            continue
        if trace:
            print '\t', '(%d)' % c, 0, -1, 0
        heapq.heappush(fringe, (0, -1, 0, [c]))
    while True:
        d, s, t, elim = heapq.heappop(fringe)
        if trace:
            print tuple(elim), d, s, t
        if len(elim) == k:
            return d
        now = time.time()
        timeout -= now - then
        if timeout <= 0.0:
            return -1
        then = now
        elim_set = set(elim)
        prefixes = candidates - elim_set
        for c in prefixes:
            reduced = root.deepcopy()
            for e in prefixes - set((c,)):
                reduced.eliminate(e)
            new_elim = [c] + elim
            if trace:
                print '\t', tuple(new_elim),
                sys.stdout.flush()
            d = ilp.distance_to(reduced, ranks, new_elim, timeout)
            if d == -1:
                return -1
            if d <= ub:
                s = -len(new_elim)
                t = len(set(new_elim) - tertiary[len(new_elim)-1])
                if trace:
                    print d, s, t
                heapq.heappush(fringe, (d, s, t, new_elim))
            elif trace:
                print d
Пример #54
0
 def pop_state(
         self, next_states: List[Tuple[int, int,
                                       PuzzleState]]) -> PuzzleState:
     return heapq.heappop(next_states)[2]
Пример #55
0
            def best_first_search(
                R, zf, K, optimal_detection, N, complexity, constellation
            ):  # best first serach 其實類似BFS search,只不過變為priority queue
                # R為H進行QR分解後的R矩陣
                # zf向量為zero forcing detection後尚未demapping的結果、K為選擇branch時要選擇離soft value最近的K個節點
                # optimal_detection向量為最終detection後得到的結果
                # N為傳送向量長度
                # complexity有3個元素,分別記錄經過幾個node、做幾次加法運算、做幾次乘法運算
                # constellation為星座圖,也就是向量中元素的值域

                priority_queue = []

                # tree的第一層時對應到的soft value為zf[N-1, 0]
                # 我們接下來要看哪些星座點離zf[N-1, 0]最近
                select = [0] * len(
                    constellation
                )  # 若select[i] == 0 代表不選擇星座點i,若select[i] == 1 代表會選擇星座點i
                lt = [[0] * 2 for i in range(len(constellation))]
                for i in range(len(constellation)):
                    lt[i][0] = i  # 代表這是第i個星座點,待會排序後可以知道到底哪個星座點與soft value最近
                    lt[i][1] = abs(zf[N - 1, 0] -
                                   constellation[i])  # 代表此星座點與這個soft value間的距離
                lt.sort(key=lambda cust: cust[1]
                        )  # 以星座點與 soft value間的距離為依據來排序,間距越小者排越前面
                # 若lt[0][0] = 3 代表星座點3與soft value最近
                # 若lt[1][0] = 1 代表星座點1與soft value第二近
                # 若lt[2][0] = 0 代表星座點0與soft value第三近

                # 注意到我們現在只要選擇K個離soft value最近的星座點
                # 所以現在開始來選擇
                count = 0
                for i in range(len(constellation)):
                    select[lt[i][0]] = 1
                    count += 1
                    if count >= K:
                        break

                # 再來將tree中第一層離soft value zf[N-1, 0]較近的星座點丟到priority queue中
                for i in range(len(constellation)):
                    if select[
                            i] == 0:  # 若星座點i 離soft value zf[N-1, 0]較遠,就不選擇此星座點了
                        continue
                    vector = np.matrix([0j] * (N)).transpose()
                    vector[N - 1, 0] = constellation[i]
                    heapq.heappush(
                        priority_queue,
                        (abs(z[N - 1, 0] - R[N - 1, N - 1] * constellation[i])
                         **2, vector, 1))
                    # 將(  (abs(z[N-1,0] - R[N-1,N-1]*constellation[i])**2, vector, 1)   ) 丟到priority queue中
                    # 其中(abs(z[N-1,0] - R[N-1,N-1]*constellation[i])**2值越小的話,優先權越大
                    # vector則存放選擇後的向量結果
                    # 1代表這個向量只有1個元素

                    complexity[0] += 1  # 經過node點數加1
                    # 做一次乘法運算時,我們會採用以下的方法,所以記錄成做一次加法
                    # 因為detect[j,0]的值域在64QAM時為 -7, -5, -3, -1, 1, 3, 5, 7
                    # 乘上1,相當於0次加法
                    # 乘上3,相當往左shift一個bit(x2)再做一次加法 = 一次加法
                    # 乘上5,相當往左shift兩個bit(x4)再做一次加法 = 一次加法
                    # 乘上7,相當往左shift三個bit(x8)再做一次減法 = 一次加法
                    if constellation[i] != 1 and constellation[i] != -1:
                        complexity[1] += 1  # 代表加法運算次數加1
                    complexity[1] += 1  # 因為有做一次減法,所以加法運算次數加1
                    complexity[2] += 1  # 因為最後有取絕對值平方,所以乘法運算次數加1

                # 接下來可以開始進行best first search了 (其過程類似BFS,只不過此處的queue變為priority queue)
                while True:
                    #先取出priority queue中優先權最高的元素
                    first_element = heapq.heappop(priority_queue)
                    # first_element[0] 存放上一層的accumulated_metric
                    # first_element[1] 存放上一層的vector
                    # first_element[2] 則代表上一層的vector有幾個元素

                    if first_element[2] == N:  # 若搜尋完畢
                        break

                    # 搜尋此節點的下層節點
                    # tree的這一層對應到的soft value為zf[N-1-first_element[2], 0]
                    # 我們接下來要看哪些星座點離zf[N-1-first_element[2], 0]最近
                    select = [0] * len(
                        constellation
                    )  # 若select[i] == 0 代表不選擇星座點i,若select[i] == 1 代表會選擇星座點i
                    lt = [[0] * 2 for i in range(len(constellation))]
                    for i in range(len(constellation)):
                        lt[i][
                            0] = i  # 代表這是第i個星座點,待會排序後可以知道到底哪個星座點與soft value最近
                        lt[i][1] = abs(
                            zf[N - 1 - first_element[2], 0] -
                            constellation[i])  # 代表此星座點與這個soft value間的距離
                    lt.sort(key=lambda cust: cust[1]
                            )  # 以星座點與 soft value間的距離為依據來排序,間距越小者排越前面
                    # 若lt[0][0] = 3 代表星座點3與soft value最近
                    # 若lt[1][0] = 1 代表星座點1與soft value第二近
                    # 若lt[2][0] = 0 代表星座點0與soft value第三近

                    # 注意到我們現在只要選擇K個離soft value最近的星座點
                    # 所以現在開始來選擇
                    count = 0
                    for i in range(len(constellation)):
                        select[lt[i][0]] = 1
                        count += 1
                        if count >= K:
                            break

                    for i in range(len(constellation)):
                        if select[
                                i] == 0:  # 若星座點i 離soft value zf[N - 1 - first_element[2], 0]較遠,就不選擇此星座點了
                            continue
                        vector = np.matrix(first_element[1])
                        vector[N - 1 - first_element[2], 0] = constellation[i]
                        complexity[0] += 1  # 經過node點數加1

                        # 接下來計算accumulated_metric
                        accumulated_metric = z[N - 1 - first_element[2], 0]
                        for j in range(N - 1, N - 2 - first_element[2], -1):
                            accumulated_metric -= R[
                                N - 1 - first_element[2],
                                j] * vector[j, 0]  # 注意到每次都有一個減法 and 乘法運算
                            # 做一次乘法運算時,我們會採用以下的方法,所以記錄成做一次加法
                            # 因為detect[j,0]的值域在64QAM時為 -7, -5, -3, -1, 1, 3, 5, 7
                            # 乘上1,相當於0次加法
                            # 乘上3,相當往左shift一個bit(x2)再做一次加法 = 一次加法
                            # 乘上5,相當往左shift兩個bit(x4)再做一次加法 = 一次加法
                            # 乘上7,相當往左shift三個bit(x8)再做一次減法 = 一次加法
                            if vector[j, 0] != 1 and vector[j, 0] != -1:
                                complexity[1] += 1  # 代表加法運算次數加1
                            complexity[1] += 1  # 因為有做一次減法,所以加法運算次數加1

                        accumulated_metric = accumulated_metric**2
                        complexity[2] += 1  # 因為做一次平方運算=一次乘法運算

                        heapq.heappush(priority_queue,
                                       (first_element[0] + accumulated_metric,
                                        vector, first_element[2] + 1))
                        # accumulated_matrix 變為first_element[0]+accumulated_metric
                        # 所以要做一次加法運算
                        complexity[1] += 1

                # first_element[1] 這個向量即為我們要的答案,將此向量的值複製到optimal_detection向量中
                for i in range(len(optimal_detection)):
                    optimal_detection[i, 0] = first_element[1][i, 0]
Пример #56
0
import heapq, sys

N = int(input())
arr = []
for i in range(N):
    a = -int(sys.stdin.readline())
    if a == 0:
        if arr:
            print(-heapq.heappop(arr))
        else:
            print(0)
    else:
        heapq.heappush(arr, a)
Пример #57
0
 def heappop(self):
     return heapq.heappop(self)
Пример #58
0
            def sphere_decoding_bfs(R, zf, K, optimal_detection, K1, N,
                                    complexity, constellation):
                # R為H進行QR分解後的R矩陣、zf向量為zero forcing detection後尚未demapping的結果、K為選擇branch時要選擇離soft value最近的K個節點
                # optimal_detection向量為最終detection後得到的結果、K1為BFS搜尋中最多准許有幾個node出現
                # N為傳送向量長度
                # complexity有3個元素,分別記錄經過幾個node、做幾次加法運算、做幾次乘法運算
                # constellation為星座圖,也就是向量中元素的值域

                # 此處的bfs比較特別,這種bfs在搜尋一層樹時,每一條節點最多只會往下選擇其中離soft value較近的K個node
                # 除此之外下一層最多總共只含K1個節點,所以我用priority queue來決定要選擇哪K1個node
                queue = [[], []]
                # 有兩個queue,a是要負責pop元素,b則push元素
                # 等到其中a  queue的元素pop完後
                # 換b queue pop元素,a  queue push元素

                current = 0  # 利用current來決定目前是哪個queue要pop元素

                # tree的第一層時對應到的soft value為zf[N-1, 0]
                # 我們接下來要看哪些星座點離zf[N-1, 0]最近
                select = [0] * len(
                    constellation
                )  # 若select[i] == 0 代表不選擇星座點i,若select[i] == 1 代表會選擇星座點i
                lt = [[0] * 2 for i in range(len(constellation))]
                for i in range(len(constellation)):
                    lt[i][0] = i  # 代表這是第i個星座點,待會排序後可以知道到底哪個星座點與soft value最近
                    lt[i][1] = abs(zf[N - 1, 0] -
                                   constellation[i])  # 代表此星座點與這個soft value間的距離
                lt.sort(key=lambda cust: cust[1]
                        )  # 以星座點與 soft value間的距離為依據來排序,間距越小者排越前面
                # 若lt[0][0] = 3 代表星座點3與soft value最近
                # 若lt[1][0] = 1 代表星座點1與soft value第二近
                # 若lt[2][0] = 0 代表星座點0與soft value第三近

                # 注意到我們現在只要選擇K個離soft value最近的星座點
                # 所以現在開始來選擇
                count = 0
                for i in range(len(constellation)):
                    select[lt[i][0]] = 1
                    count += 1
                    if count >= K:
                        break

                # 一開始將tree中第一層離soft value zf[N-1, 0]較近的節點丟到queue[current]中
                for i in range(len(constellation)):
                    if select[i] == 0:  # 若此點離soft value太遠就不選取
                        continue
                    vector = np.matrix([0j] * (N)).transpose()
                    vector[N - 1, 0] = constellation[i]
                    heapq.heappush(
                        queue[current],
                        (abs(z[N - 1, 0] - R[N - 1, N - 1] * constellation[i])
                         **2, vector, 1))
                    # 將(  (abs(z[N-1,0] - R[N-1,N-1]*constellation[i])**2, vector, 1)   ) 丟到priority queue中
                    # 其中(abs(z[N-1,0] - R[N-1,N-1]*constellation[i])**2值越小的話,優先權越大
                    # vector則存放選擇後的向量結果
                    # 1代表這個向量只有1個元素

                    complexity[0] += 1  # 經過node點數加1
                    # 做一次乘法運算時,我們會採用以下的方法,所以記錄成做一次加法
                    # 因為detect[j,0]的值域在64QAM時為 -7, -5, -3, -1, 1, 3, 5, 7
                    # 乘上1,相當於0次加法
                    # 乘上3,相當往左shift一個bit(x2)再做一次加法 = 一次加法
                    # 乘上5,相當往左shift兩個bit(x4)再做一次加法 = 一次加法
                    # 乘上7,相當往左shift三個bit(x8)再做一次減法 = 一次加法
                    if constellation[i] != 1 and constellation[i] != -1:
                        complexity[1] += 1  # 代表加法運算次數加1
                    complexity[1] += 1  # 因為有做一次減法,所以加法運算次數加1
                    complexity[2] += 1  # 因為最後有取絕對值平方,所以乘法運算次數加1

                # 接下來要開始將queue[current]的元素pop到另一個queue[(current+1)%2]中
                while True:

                    count = 0  # count用來紀錄queue[current] 目前pop幾個元素
                    while True:  # 將queue[current]的元素pop出來,並根據此pop出來的元素從tree的下層選出其他節點加到queue[(current+1)%2]中
                        # 注意到queue[current]最多只會pop K個node

                        if len(queue[current]
                               ) == 0:  # 若該queue的元素都pop出來了,就直接break
                            break
                        elif count >= K1:  # 若已經從queue[current]中pop K1個元素(代表從這一層的節點選取K1個節點了)
                            # 將queue[current]清空後在break
                            while True:
                                heapq.heappop(queue[current])
                                if len(queue[current]) == 0:
                                    break
                            break

                        # 先取出priority queue中優先權最高的元素
                        first_element = heapq.heappop(queue[current])
                        # first_element[0] 存放上一層的accumulated_metric
                        # first_element[1] 存放上一層的vector
                        # first_element[2] 則代表上一層的vector有幾個元素

                        count += 1  # 因為queue[current] pop出一個元素了

                        if first_element[2] == N:  # 代表BFS已搜尋到樹的最底端,搜尋結束
                            break

                        # 接下來搜尋此節點的下層節點
                        # 搜尋前我們要看,哪一個下層節點離下層soft value值較近,我們只選取離soft value值較近的K個節點
                        # 我們接下來要看哪些星座點離zf[N - 1 - first_element[2], 0]最近
                        select = [0] * len(
                            constellation
                        )  # 若select[i] == 0 代表不選擇星座點i,若select[i] == 1 代表會選擇星座點i
                        lt = [[0] * 2 for i in range(len(constellation))]
                        for i in range(len(constellation)):
                            lt[i][
                                0] = i  # 代表這是第i個星座點,待會排序後可以知道到底哪個星座點與soft value最近
                            lt[i][1] = abs(
                                zf[N - 1 - first_element[2], 0] -
                                constellation[i])  # 代表此星座點與這個soft value間的距離
                        lt.sort(key=lambda cust: cust[1]
                                )  # 以星座點與 soft value間的距離為依據來排序,間距越小者排越前面
                        # 若lt[0][0] = 3 代表星座點3與soft value最近
                        # 若lt[1][0] = 1 代表星座點1與soft value第二近
                        # 若lt[2][0] = 0 代表星座點0與soft value第三近

                        # 注意到我們現在只要選擇K個離soft value最近的星座點
                        # 所以現在開始來選擇
                        count1 = 0
                        for i in range(len(constellation)):
                            select[lt[i][0]] = 1
                            count1 += 1
                            if count1 >= K:
                                break

                        for i in range(len(constellation)):
                            if select[i] == 0:  # 若該星座點離此層的soft value值較遠就不選取
                                continue
                            vector = np.matrix(first_element[1])
                            vector[N - 1 - first_element[2],
                                   0] = constellation[i]
                            complexity[0] += 1  # 經過node點數加1

                            # 接下來計算accumulated_metric
                            accumulated_metric = z[N - 1 - first_element[2], 0]
                            for j in range(N - 1, N - 2 - first_element[2],
                                           -1):
                                accumulated_metric -= R[
                                    N - 1 - first_element[2],
                                    j] * vector[j, 0]  # 注意到每次都有一個減法 and 乘法運算
                                # 做一次乘法運算時,我們會採用以下的方法,所以記錄成做一次加法
                                # 因為detect[j,0]的值域在64QAM時為 -7, -5, -3, -1, 1, 3, 5, 7
                                # 乘上1,相當於0次加法
                                # 乘上3,相當往左shift一個bit(x2)再做一次加法 = 一次加法
                                # 乘上5,相當往左shift兩個bit(x4)再做一次加法 = 一次加法
                                # 乘上7,相當往左shift三個bit(x8)再做一次減法 = 一次加法
                                if vector[j, 0] != 1 and vector[j, 0] != -1:
                                    complexity[1] += 1  # 代表加法運算次數加1
                                complexity[1] += 1  # 因為有做一次減法,所以加法運算次數加1

                            accumulated_metric = accumulated_metric**2
                            complexity[2] += 1  # 因為做一次平方運算=一次乘法運算

                            heapq.heappush(
                                queue[(current + 1) % 2],
                                (first_element[0] + accumulated_metric, vector,
                                 first_element[2] +
                                 1))  # 將一個節點push到queue[(current+1)%2]中
                            # accumulated_matrix 變為first_element[0]+accumulated_metric
                            # 所以要做一次加法運算
                            complexity[1] += 1

                    current = (
                        current + 1
                    ) % 2  # 因為待會負責pop的queue會變成要push、而剛剛負責push的queue則要負責pop

                    if first_element[2] == N:
                        break

                # first_element[1] 這個向量即為我們要的答案,將此向量的值複製到optimal_detection向量中
                for i in range(len(optimal_detection)):
                    optimal_detection[i, 0] = first_element[1][i, 0]
Пример #59
0
visited = [False]*(n+1)

edge = [[3, 6], [4, 3], [5, 2], [1, 3], [1, 5], [3, 5], [5, 6]]

for a, b in edge:
    graph[a].append((a, b))
    graph[b].append((b, a))

result[1] = 0
for i in range(len(graph[1])):
    heapq.heappush(heapq_list, graph[1][i])
    visited[1] = True

while(heapq_list):
    print(heapq_list)
    from_, to_ = heapq.heappop(heapq_list)
    if visited[to_] == True:
        continue
    if result[from_]+1 < result[to_]:
        result[to_] = result[from_]+1
    for i in range(len(graph[to_])):
        heapq.heappush(heapq_list, graph[to_][i])
    visited[to_] = True

max_num = max(result)
count = 0
for num in result:
    if num == max_num:
        count += 1
print(result)
print(count)
Пример #60
0
def heap_sort(arr):
	heapq.heapify(arr)  # O(n) time
	result = []
	while len(arr) > 0:
		result.append(heapq.heappop(arr))  # Pop takes O(1) + O(log n) for re-heapification
	return result