python
class PossibleBipartition:
"""
可能的二分法
「其实考察的就是二分图的判定」
用dfs和bfs 两种方法分别实现
https://leetcode.cn/problems/possible-bipartition/
"""
def __init__(self):
self.success = True
self.color = []
self.visited = []
def dfs(self, n, dislikes):
"""
DFS递归实现
:param n:
:param dislikes:
:return:
"""
# 图节点编号为 1...n
self.color = [False] * (n+1)
self.visited = [False] * (n+1)
graph = self.buildgraph(n, dislikes)
# 因为图不一定是联通的，可能存在多个子图
# 所以要把每个节点都作为起点进行一次遍历
# 如果发现任何一个子图不是二分图，整幅图都不是二分图
for v in range(1, n+1):
if not self.visited[v]:
self.dfs_traverse(graph, v)
return self.success
def buildgraph(self, n, dislikes):
graph = [[] for _ in range(n+1)]
for edge in dislikes:
v = edge[1]
w = edge[0]
# 无向图相当于双向图
graph[v].append(w)
graph[w].append(v)
return graph
def dfs_traverse(self, graph, v):
if not self.success:
return
self.visited[v] = True
for w in graph[v]:
if not self.visited[w]:
self.color[w] = not self.color[v]
self.dfs_traverse(graph, w)
else:
if self.color[v] == self.color[w]:
self.success = False
return
def bfs(self, n, dislikes):
"""
BFS实现，用队列替代递归调用
:param n:
:param dislikes:
:return:
"""
# 图节点编号为 1...n
self.color = [False] * (n + 1)
self.visited = [False] * (n + 1)
graph = self.buildgraph(n, dislikes)
# 因为图不一定是联通的，可能存在多个子图
# 所以要把每个节点都作为起点进行一次遍历
# 如果发现任何一个子图不是二分图，整幅图都不是二分图
for v in range(1, n + 1):
if not self.visited[v]:
self.bfs_traverse(graph, v)
return self.success
def bfs_traverse(self, graph, start):
# 节点队列
queue = []
self.visited[start] = True
queue.append(start)
while queue and self.success:
v = queue.pop(0)
# 从节点 v 向所有相邻节点扩散
for w in graph[v]:
if not self.visited[w]:
# 相邻节点w没有被访问过
# 那么应该给节点w涂上和节点v不同的颜⾊
self.color[w] = not self.color[v]
# 标记 w 节点，并放⼊队列
self.visited[w] = True
queue.append(w)
else:
if self.color[v] == self.color[w]:
self.success = False
return