def main():
print "\ncheck stack"
stack = Stack(1, 2, 34, 5)
for x in range(0, 5):
stack.push(x)
print stack
for x in range(0, 15):
print "".join(["size=", str(len(stack)), " cur_node=", str(stack.pop())])
print "\ncheck queue"
queue = Queue(1, 2, 34, 5)
for x in range(0, 5):
queue.enter(x)
print stack
for x in range(0, 15):
print "".join(["size=", str(len(queue)), " cur_node=", str(queue.exit())])
print "\ncheck BSTree"
tree = BSTree(1, 2, 34, 5)
print tree
print tree.find(10)
print tree.find(5)
print tree.max()
print tree.min()
print tree.successor(34)
print tree.successor(5)
print tree.predecessor(1)
print tree.predecessor(2)
def flatorder(self, f):
result = []
queue = Queue(self.__root)
while not queue.empty():
cur_node = queue.exit()
result.append(f(cur_node.value))
if cur_node.left is not None:
queue.enter(cur_node.left)
if cur_node.right is not None:
queue.enter(cur_node.right)
return result
def print_tree(self):
"""
打印树的结构
:return:
"""
queue = Queue(self.__root)
next_level = 1
now_node_count = 0
while not queue.empty():
cur_node = queue.exit()
print str(cur_node) + "\t",
now_node_count += 1
if now_node_count == next_level:
print
now_node_count = 0
next_level *= 2
if cur_node.left is not None:
queue.enter(cur_node.left)
if cur_node.right is not None:
queue.enter(cur_node.right)
def print_tree(self):
"""
打印树的结构
:return:
"""
queue = Queue(self.__root)
next_level = 1
now_node_count = 0
while not queue.empty():
cur_node = queue.exit()
print str(cur_node) + "\t",
now_node_count += 1
if now_node_count == next_level:
print
now_node_count = 0
next_level *= 2
if cur_node.left is not None:
queue.enter(cur_node.left)
if cur_node.right is not None:
queue.enter(cur_node.right)
def lws(self, from_node, to_node):
"""
在有向无环带权图里面查找最长路径
令list(s,t)是从s到t的最长带权路径。
那么可以使用递归式表示这个问题
list(s,t) = max(list(s,t-1))+list(t-1,t)(唯一)
用动态规划自下而上解决，因为自上而下解决首先遇到的问题就是
查找指向一个节点的节点在邻接表中比较困难
:param from_node:
:param to_node:
:return:
"""
__lws = {}
# 为了计算方便，这里把开始节点到开始节点插入字典中，
zero_edge = Edge(from_node, from_node)
__lws[zero_edge] = 0
graph_stack = Queue()
graph_stack.enter(from_node)
while not graph_stack.empty():
cur_node = graph_stack.exit()
cur_edge_list = self.__adj_list.get(cur_node)
if cur_edge_list is None:
print ",".join(map(lambda edge: str(edge), __lws.iteritems()))
print ",".join(map(lambda edge: str(edge), __lws))
return __lws[Edge(from_node, to_node)]
for edge_end_node in cur_edge_list:
graph_stack.enter(edge_end_node.key)
last_weighted_length = __lws[Edge(from_node, cur_node)]
cur_edge = Edge(from_node, edge_end_node.key)
cur_weight_length = last_weighted_length + edge_end_node.weight
# 如果不存在这个边，那么就插入
if cur_edge not in __lws:
__lws[cur_edge] = cur_weight_length
# 如果存在，那么就把最大值插入
elif cur_weight_length > __lws[cur_edge]:
__lws[cur_edge] = cur_weight_length
print ",".join(map(lambda edge: str(edge), __lws.iteritems()))
print ",".join(map(lambda edge: str(edge), __lws))
return __lws[Edge(from_node, to_node)]
def lws(self, from_node, to_node):
"""
在有向无环带权图里面查找最长路径
令list(s,t)是从s到t的最长带权路径。
那么可以使用递归式表示这个问题
list(s,t) = max(list(s,t-1))+list(t-1,t)(唯一)
用动态规划自下而上解决，因为自上而下解决首先遇到的问题就是
查找指向一个节点的节点在邻接表中比较困难
:param from_node:
:param to_node:
:return:
"""
__lws = {}
# 为了计算方便，这里把开始节点到开始节点插入字典中，
zero_edge = Edge(from_node, from_node)
__lws[zero_edge] = 0
graph_stack = Queue()
graph_stack.enter(from_node)
while not graph_stack.empty():
cur_node = graph_stack.exit()
cur_edge_list = self.__adj_list.get(cur_node)
if cur_edge_list is None:
print ",".join(map(lambda edge: str(edge), __lws.iteritems()))
print ",".join(map(lambda edge: str(edge), __lws))
return __lws[Edge(from_node, to_node)]
for edge_end_node in cur_edge_list:
graph_stack.enter(edge_end_node.key)
last_weighted_length = __lws[Edge(from_node, cur_node)]
cur_edge = Edge(from_node, edge_end_node.key)
cur_weight_length = last_weighted_length + edge_end_node.weight
# 如果不存在这个边，那么就插入
if cur_edge not in __lws:
__lws[cur_edge] = cur_weight_length
# 如果存在，那么就把最大值插入
elif cur_weight_length > __lws[cur_edge]:
__lws[cur_edge] = cur_weight_length
print ",".join(map(lambda edge: str(edge), __lws.iteritems()))
print ",".join(map(lambda edge: str(edge), __lws))
return __lws[Edge(from_node, to_node)]