Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.
Each person may dislike some other people, and they should not go into the same group.
Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.
Return true if and only if it is possible to split everyone into two groups in this way.
Example 1:
Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4], group2 [2,3]
Example 2:
Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false
Example 3:
Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false
Solution
我们把N个人作为图的顶点，不喜欢关系作为边，构造出一个不喜欢的图。然后进行DFS遍历，同时给每个人黑白两色，相邻的不能有相同的颜色。如果冲突，则返回False，否则如果遍历完整个图没发现冲突则返回True。
class Solution:
def possibleBipartition(self, N: int, dislikes: List[List[int]]) -> bool:
edges = [[] for i in range(N + 1)]
colors = [0] * (N + 1)
for u,v in dislikes:
edges[u].append(v)
edges[v].append(u)
for i in range(1,N+1):
if colors[i]==0:
q=[i]
colors[i] = 1
while q:
cur = q.pop()
cur_c =colors[cur]
for edge in edges[cur]:
if colors[edge]==0:
q.append(edge)
colors[edge] = -1 *cur_c
elif (colors[edge] ==cur_c):
return False
return True
