def solve(file_name):
graph = dijstra.read_input(file_name)
dist = dijstra.dijstra(graph, 1)
result = [0] * len(dist)
for n in dist:
result[n-1] = dist[n]
return result
def johnson(graph):
"""Given a graph, find shortest distance between all pairs."""
# 1. Run bellman-ford to get node shift value
# assume a artifical node, with edge cost 0 to all other nodes
mat = {}
for n in graph:
mat[n] = 0
changed = True
round = 1
while changed and round <= len(graph):
changed = False
round += 1
for es in graph:
for et in graph[es]:
edge_cost = graph[es][et]
if mat[et] > mat[es] + edge_cost:
changed = True
mat[et] = mat[es] + edge_cost
if changed:
return 'negative cycle'
# 2. Build a graph with edge cost shifted as:
# new_cost = old_cost + mat[start] - mat[end]
new_graph = {}
for es in graph:
new_graph[es] = {}
for et in graph[es]:
new_graph[es][et] = graph[es][et] + mat[es] - mat[et]
assert new_graph[es][et] >= 0
# 3. Run dijstra n times
result = []
for es in graph:
costs = dijstra(new_graph, es)
for et, dist in costs:
result.append((es, et, dist - mat[es] + mat[et]))
return result
def johnson(graph):
"""Given a graph, find shortest distance between all pairs."""
# 1. Run bellman-ford to get node shift value
# assume a artifical node, with edge cost 0 to all other nodes
mat = {}
for n in graph:
mat[n] = 0
changed = True
round = 1
while changed and round <= len(graph):
changed = False
round += 1
for es in graph:
for et in graph[es]:
edge_cost = graph[es][et]
if mat[et] > mat[es] + edge_cost:
changed = True
mat[et] = mat[es] + edge_cost
if changed:
return 'negative cycle'
# 2. Build a graph with edge cost shifted as:
# new_cost = old_cost + mat[start] - mat[end]
new_graph = {}
for es in graph:
new_graph[es] = {}
for et in graph[es]:
new_graph[es][et] = graph[es][et] + mat[es] - mat[et]
assert new_graph[es][et] >= 0
# 3. Run dijstra n times
result = []
for es in graph:
costs = dijstra(new_graph, es)
for et, dist in costs:
result.append((es, et, dist - mat[es] + mat[et]))
return result
