[英]Find the shortest cycle in a graph (Undirected, Unweighted)
I've been trying to write a python program that finds the shortest cycle in a graph, but I'm stuck.我一直在尝试编写一个 python 程序,它可以在图中找到最短的循环,但我被卡住了。 This is my code:这是我的代码:
def shortestCycle(vertices):
distances = []
for i in range(len(vertices.keys())):
dist = 0
visited = []
vertex = list(vertices.keys())[i]
searchQueue = deque()
searchQueue += vertices[vertex]
while searchQueue:
currentVertex = searchQueue.popleft()
if currentVertex not in visited:
if currentVertex == vertex:
break
else:
visited.append(currentVertex)
searchQueue += vertices[currentVertex]
dist += 1
distances.append(udaljenost)
return min(distances)
For some extra explanation, vertices
is a dictionary with vertices as keys, and their neighbours as values, for example: {1 : [2, 4, 5, 8], 2 : [1, 3], ...
.对于一些额外的解释, vertices
是一个字典,以顶点为键,它们的邻居为值,例如: {1 : [2, 4, 5, 8], 2 : [1, 3], ...
。 This code gives the wrong result, and I'm partialy aware of what is causing that, but I'm not sure how to fix it.这段代码给出了错误的结果,我部分知道是什么原因造成的,但我不确定如何解决它。 Any ideas or advices ?任何想法或建议?
Edit:编辑:
example input: {'1': ['2', '4', '5', '8'], '2': ['1', '3'], '3': ['2', '4'], '4': ['1', '3'], '5': ['1', '6'], '6': ['5', '7'], '7': ['6', '8'], '8': ['1', '7']}
示例输入: {'1': ['2', '4', '5', '8'], '2': ['1', '3'], '3': ['2', '4'], '4': ['1', '3'], '5': ['1', '6'], '6': ['5', '7'], '7': ['6', '8'], '8': ['1', '7']}
example output: 4
示例输出: 4
I managed to solve it:我设法解决了它:
def shortestCycle(vertices):
minCycle = sys.maxsize
n = len(vertices.keys())
for i in range(n):
distances = [sys.maxsize for i in range(n)]
parent = [-1 for i in range(n)]
distances[i] = 0
q = deque()
q.append(i + 1)
while q:
currentVertex = str(q.popleft())
for v in vertices[currentVertex]:
j = currentVertex
if distances[v - 1] == sys.maxsize:
distances[v - 1] = 1 + distances[j - 1]
parent[v - 1] = j
q.append(v)
elif parent[j - 1] != v and parent[v - 1] != j - 1:
minCycle = min(minCycle, distances[j - 1] + distances[v - 1] + 1)
if minCycle == sys.maxsize:
return -1
else:
return minCycle
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