# 并查集模板
class UnionFind:
def __init__(self, n: int):
self.parent = list(range(n))
self.size = [1] * n
self.n = n
# 当前连通分量数目
self.setCount = n
def findset(self, x: int) -> int:
if self.parent[x] == x:
return x
self.parent[x] = self.findset(self.parent[x])
return self.parent[x]
def unite(self, x: int, y: int) -> bool:
x, y = self.findset(x), self.findset(y)
if x == y:
return False
if self.size[x] < self.size[y]:
x, y = y, x
self.parent[y] = x
self.size[x] += self.size[y]
self.setCount -= 1
return True
def connected(self, x: int, y: int) -> bool:
x, y = self.findset(x), self.findset(y)
return x == y
class Solution:
def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
m = len(edges)
for i, edge in enumerate(edges):
edge.append(i)
edges.sort(key=lambda x: x[2])
# 计算 value
uf_std = UnionFind(n)
value = 0
for i in range(m):
if uf_std.unite(edges[i][0], edges[i][1]):
value += edges[i][2]
ans = [list(), list()]
for i in range(m):
# 判断是否是关键边
uf = UnionFind(n)
v = 0
for j in range(m):
if i != j and uf.unite(edges[j][0], edges[j][1]):
v += edges[j][2]
if uf.setCount != 1 or (uf.setCount == 1 and v > value):
ans[0].append(edges[i][3])
continue
# 判断是否是伪关键边
uf = UnionFind(n)
uf.unite(edges[i][0], edges[i][1])
v = edges[i][2]
for j in range(m):
if i != j and uf.unite(edges[j][0], edges[j][1]):
v += edges[j][2]
if v == value:
ans[1].append(edges[i][3])
return ans
# 并查集,困难,每日一题
# https://leetcode-cn.com/problems/find-critical-and-pseudo-critical-edges-in-minimum-spanning-tree/