我试图转换DFS程序以检查图形是否为二分图。我想通过路径并将访问过的节点发送到不同的子集,只要它们不相邻。例如:
如果路径如下所示:1-> 2-> 3-> 4-> 5
两个子集应如下所示:
[1,3,5] [2,4]
这是DFS代码:
def dfs(x, node, visited):
if node not in visited:
visited.append(node)
for n in x[node]:
dfs(x,n, visited)
return visited
答案 0 :(得分:0)
测试二分法是通过"着色"当您执行DFS时,相邻节点具有交替的颜色,如果任何两个节点使用相同的颜色"颜色"然后图表不是二分图。您可以通过在执行DFS时将节点放入两个不同的集合来执行此操作:
def bipart(x, node, visited, set1, set2):
if node in set2: # if a node has already been put into the other set, the graph is not bipartite
return False
if node not in visited: # if we haven't seen this yet
visited.add(node) # mark it as visited
set1.add(node) # put it in the first set
for n in x[node]: # for each neighbor
bipart(x, n, set2, set1) # put it into the other set
return True # if we get here we have separated the graph into two sets where all the neighbors of a node in one set are in the other set.
请注意,我将visited
设置为一个集合而不是列表,因为检查集合的成员资格会更快。