1489. 找到最小生成树里的关键边和伪关键边
给你一个 n
个点的带权无向连通图,节点编号为 0
到 n-1
,同时还有一个数组 edges
,其中 edges[i] = [from
i, toi, weighti]
表示在 fromi
和 toi
节点之间有一条带权无向边。最小生成树 (MST) 是给定图中边的一个子集,它连接了所有节点且没有环,而且这些边的权值和最小。
请你找到给定图中最小生成树的所有关键边和伪关键边。如果从图中删去某条边,会导致最小生成树的权值和增加,那么我们就说它是一条关键边。伪关键边则是可能会出现在某些最小生成树中但不会出现在所有最小生成树中的边。
请注意,你可以分别以任意顺序返回关键边的下标和伪关键边的下标。
示例 1:
输入:n = 5, edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]] 输出:[[0,1],[2,3,4,5]] 解释:上图描述了给定图。 下图是所有的最小生成树。 注意到第 0 条边和第 1 条边出现在了所有最小生成树中,所以它们是关键边,我们将这两个下标作为输出的第一个列表。 边 2,3,4 和 5 是所有 MST 的剩余边,所以它们是伪关键边。我们将它们作为输出的第二个列表。
示例 2 :
输入:n = 4, edges = [[0,1,1],[1,2,1],[2,3,1],[0,3,1]] 输出:[[],[0,1,2,3]] 解释:可以观察到 4 条边都有相同的权值,任选它们中的 3 条可以形成一棵 MST 。所以 4 条边都是伪关键边。
提示:
2 <= n <= 100
1 <= edges.length <= min(200, n * (n - 1) / 2)
edges[i].length == 3
0 <= fromi < toi < n
1 <= weighti <= 1000
(fromi, toi)
数对都是互不相同的。原站题解
javascript 解法, 执行用时: 180 ms, 内存消耗: 51.7 MB, 提交时间: 2023-10-10 15:22:54
/** * @param {number} n * @param {number[][]} edges * @return {number[][]} */ var findCriticalAndPseudoCriticalEdges = function(n, edges) { const m = edges.length; for (const [i, edge] of edges.entries()) { edge.push(i); } edges.sort((a, b) => a[2] - b[2]); // 计算 value const uf_std = new UnionFind(n); let value = 0; for (let i = 0; i < m; i++) { if (uf_std.unite(edges[i][0], edges[i][1])) { value += edges[i][2]; } } const ans = [[], []]; for (let i = 0; i < m; i++) { // 判断是否是关键边 let uf = new UnionFind(n); let v = 0; for (let j = 0; j < m; j++) { if (i !== j && uf.unite(edges[j][0], edges[j][1])) { v += edges[j][2]; } } if (uf.setCount !== 1 || (uf.setCount === 1 && v > value)) { ans[0].push(edges[i][3]); continue; } // 判断是否是伪关键边 uf = new UnionFind(n); uf.unite(edges[i][0], edges[i][1]); v = edges[i][2]; for (let j = 0; j < m; j++) { if (i !== j && uf.unite(edges[j][0], edges[j][1])) { v += edges[j][2]; } } if (v === value) { ans[1].push(edges[i][3]); } } return ans; }; // 并查集模板 class UnionFind { constructor (n) { this.parent = new Array(n).fill(0).map((element, index) => index); this.size = new Array(n).fill(1); // 当前连通分量数目 this.setCount = n; } findset (x) { if (this.parent[x] === x) { return x; } this.parent[x] = this.findset(this.parent[x]); return this.parent[x]; } unite (a, b) { let x = this.findset(a), y = this.findset(b); if (x === y) { return false; } if (this.size[x] < this.size[y]) { [x, y] = [y, x]; } this.parent[y] = x; this.size[x] += this.size[y]; this.setCount -= 1; return true; } connected (a, b) { const x = this.findset(a), y = this.findset(b); return x === y; } }
python3 解法, 执行用时: 1268 ms, 内存消耗: 16.2 MB, 提交时间: 2023-10-10 15:22:30
# 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
java 解法, 执行用时: 24 ms, 内存消耗: 43 MB, 提交时间: 2023-10-10 15:22:10
class Solution { public List<List<Integer>> findCriticalAndPseudoCriticalEdges(int n, int[][] edges) { int m = edges.length; int[][] newEdges = new int[m][4]; for (int i = 0; i < m; ++i) { for (int j = 0; j < 3; ++j) { newEdges[i][j] = edges[i][j]; } newEdges[i][3] = i; } Arrays.sort(newEdges, new Comparator<int[]>() { public int compare(int[] u, int[] v) { return u[2] - v[2]; } }); // 计算 value UnionFind ufStd = new UnionFind(n); int value = 0; for (int i = 0; i < m; ++i) { if (ufStd.unite(newEdges[i][0], newEdges[i][1])) { value += newEdges[i][2]; } } List<List<Integer>> ans = new ArrayList<List<Integer>>(); for (int i = 0; i < 2; ++i) { ans.add(new ArrayList<Integer>()); } for (int i = 0; i < m; ++i) { // 判断是否是关键边 UnionFind uf = new UnionFind(n); int v = 0; for (int j = 0; j < m; ++j) { if (i != j && uf.unite(newEdges[j][0], newEdges[j][1])) { v += newEdges[j][2]; } } if (uf.setCount != 1 || (uf.setCount == 1 && v > value)) { ans.get(0).add(newEdges[i][3]); continue; } // 判断是否是伪关键边 uf = new UnionFind(n); uf.unite(newEdges[i][0], newEdges[i][1]); v = newEdges[i][2]; for (int j = 0; j < m; ++j) { if (i != j && uf.unite(newEdges[j][0], newEdges[j][1])) { v += newEdges[j][2]; } } if (v == value) { ans.get(1).add(newEdges[i][3]); } } return ans; } } // 并查集模板 class UnionFind { int[] parent; int[] size; int n; // 当前连通分量数目 int setCount; public UnionFind(int n) { this.n = n; this.setCount = n; this.parent = new int[n]; this.size = new int[n]; Arrays.fill(size, 1); for (int i = 0; i < n; ++i) { parent[i] = i; } } public int findset(int x) { return parent[x] == x ? x : (parent[x] = findset(parent[x])); } public boolean unite(int x, int y) { x = findset(x); y = findset(y); if (x == y) { return false; } if (size[x] < size[y]) { int temp = x; x = y; y = temp; } parent[y] = x; size[x] += size[y]; --setCount; return true; } public boolean connected(int x, int y) { x = findset(x); y = findset(y); return x == y; } }
golang 解法, 执行用时: 32 ms, 内存消耗: 6.9 MB, 提交时间: 2023-10-10 15:21:56
type unionFind struct { parent, size []int setCount int // 当前连通分量数目 } func newUnionFind(n int) *unionFind { parent := make([]int, n) size := make([]int, n) for i := range parent { parent[i] = i size[i] = 1 } return &unionFind{parent, size, n} } func (uf *unionFind) find(x int) int { if uf.parent[x] != x { uf.parent[x] = uf.find(uf.parent[x]) } return uf.parent[x] } func (uf *unionFind) union(x, y int) bool { fx, fy := uf.find(x), uf.find(y) if fx == fy { return false } if uf.size[fx] < uf.size[fy] { fx, fy = fy, fx } uf.size[fx] += uf.size[fy] uf.parent[fy] = fx uf.setCount-- return true } func findCriticalAndPseudoCriticalEdges(n int, edges [][]int) [][]int { for i, e := range edges { edges[i] = append(e, i) } sort.Slice(edges, func(i, j int) bool { return edges[i][2] < edges[j][2] }) calcMST := func(uf *unionFind, ignoreID int) (mstValue int) { for i, e := range edges { if i != ignoreID && uf.union(e[0], e[1]) { mstValue += e[2] } } if uf.setCount > 1 { return math.MaxInt64 } return } mstValue := calcMST(newUnionFind(n), -1) var keyEdges, pseudokeyEdges []int for i, e := range edges { // 是否为关键边 if calcMST(newUnionFind(n), i) > mstValue { keyEdges = append(keyEdges, e[3]) continue } // 是否为伪关键边 uf := newUnionFind(n) uf.union(e[0], e[1]) if e[2]+calcMST(uf, i) == mstValue { pseudokeyEdges = append(pseudokeyEdges, e[3]) } } return [][]int{keyEdges, pseudokeyEdges} }
golang 解法, 执行用时: 4 ms, 内存消耗: 4.1 MB, 提交时间: 2023-10-10 15:21:47
type unionFind struct { parent, size []int } func newUnionFind(n int) *unionFind { parent := make([]int, n) size := make([]int, n) for i := range parent { parent[i] = i size[i] = 1 } return &unionFind{parent, size} } func (uf *unionFind) find(x int) int { if uf.parent[x] != x { uf.parent[x] = uf.find(uf.parent[x]) } return uf.parent[x] } func (uf *unionFind) union(x, y int) bool { fx, fy := uf.find(x), uf.find(y) if fx == fy { return false } if uf.size[fx] < uf.size[fy] { fx, fy = fy, fx } uf.size[fx] += uf.size[fy] uf.parent[fy] = fx return true } func findCriticalAndPseudoCriticalEdges(n int, edges [][]int) [][]int { m := len(edges) edgeType := make([]int, m) // -1:不在最小生成树中;0:伪关键边;1:关键边 for i, e := range edges { edges[i] = append(e, i) } sort.Slice(edges, func(i, j int) bool { return edges[i][2] < edges[j][2] }) type neighbor struct{ to, edgeID int } graph := make([][]neighbor, n) dfn := make([]int, n) // 遍历到该顶点时的时间戳 timestamp := 0 var tarjan func(int, int) int tarjan = func(v, pid int) int { timestamp++ dfn[v] = timestamp lowV := timestamp for _, e := range graph[v] { if w := e.to; dfn[w] == 0 { lowW := tarjan(w, e.edgeID) if lowW > dfn[v] { edgeType[e.edgeID] = 1 } lowV = min(lowV, lowW) } else if e.edgeID != pid { lowV = min(lowV, dfn[w]) } } return lowV } uf := newUnionFind(n) for i := 0; i < m; { vs := []int{} // 将权值相同的边分为一组,建图,然后用 Tarjan 算法找桥边 for weight := edges[i][2]; i < m && edges[i][2] == weight; i++ { e := edges[i] v, w, edgeID := uf.find(e[0]), uf.find(e[1]), e[3] if v != w { graph[v] = append(graph[v], neighbor{w, edgeID}) graph[w] = append(graph[w], neighbor{v, edgeID}) vs = append(vs, v, w) // 记录图中顶点 } else { edgeType[edgeID] = -1 } } for _, v := range vs { if dfn[v] == 0 { tarjan(v, -1) } } // 合并顶点、重置数据 for j := 0; j < len(vs); j += 2 { v, w := vs[j], vs[j+1] uf.union(v, w) graph[v] = nil graph[w] = nil dfn[v] = 0 dfn[w] = 0 } } var keyEdges, pseudokeyEdges []int for i, tp := range edgeType { if tp == 0 { pseudokeyEdges = append(pseudokeyEdges, i) } else if tp == 1 { keyEdges = append(keyEdges, i) } } return [][]int{keyEdges, pseudokeyEdges} } func min(a, b int) int { if a < b { return a } return b }
python3 解法, 执行用时: 68 ms, 内存消耗: 16.1 MB, 提交时间: 2023-10-10 15:21:31
# 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 # Tarjan 算法求桥边模版 class TarjanSCC: def __init__(self, n: int, edges: List[List[int]], edgesId: List[List[int]]): self.n = n self.edges = edges self.edgesId = edgesId self.low = [-1] * n self.dfn = [-1] * n self.ans = list() self.ts = -1 def getCuttingEdge(self) -> List[int]: for i in range(self.n): if self.dfn[i] == -1: self.pGetCuttingEdge(i, -1) return self.ans def pGetCuttingEdge(self, u: int, parentEdgeId: int): self.ts += 1 self.low[u] = self.dfn[u] = self.ts for v, iden in zip(self.edges[u], self.edgesId[u]): if self.dfn[v] == -1: self.pGetCuttingEdge(v, iden) self.low[u] = min(self.low[u], self.low[v]) if self.low[v] > self.dfn[u]: self.ans.append(iden) elif iden != parentEdgeId: self.low[u] = min(self.low[u], self.dfn[v]) 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]) uf = UnionFind(n) ans0 = list() label = [0] * m i = 0 while i < m: # 找出所有权值为 w 的边,下标范围为 [i, j) w = edges[i][2] j = i while j < m and edges[j][2] == edges[i][2]: j += 1 # 存储每个连通分量在图 G 中的编号 compToId = dict() # 图 G 的节点数 gn = 0 for k in range(i, j): x = uf.findset(edges[k][0]) y = uf.findset(edges[k][1]) if x != y: if x not in compToId: compToId[x] = gn gn += 1 if y not in compToId: compToId[y] = gn gn += 1 else: # 将自环边标记为 -1 label[edges[k][3]] = -1 # 图 G 的边 gm = collections.defaultdict(list) gmid = collections.defaultdict(list) for k in range(i, j): x = uf.findset(edges[k][0]) y = uf.findset(edges[k][1]) if x != y: idx, idy = compToId[x], compToId[y] gm[idx].append(idy) gmid[idx].append(edges[k][3]) gm[idy].append(idx) gmid[idy].append(edges[k][3]) bridges = TarjanSCC(gn, gm, gmid).getCuttingEdge() # 将桥边(关键边)标记为 1 ans0.extend(bridges) for iden in bridges: label[iden] = 1 for k in range(i, j): uf.unite(edges[k][0], edges[k][1]) i = j # 未标记的边即为非桥边(伪关键边) ans1 = [i for i in range(m) if label[i] == 0] return [ans0, ans1]
java 解法, 执行用时: 10 ms, 内存消耗: 42.5 MB, 提交时间: 2023-10-10 15:21:13
class Solution { public List<List<Integer>> findCriticalAndPseudoCriticalEdges(int n, int[][] edges) { int m = edges.length; int[][] newEdges = new int[m][4]; for (int i = 0; i < m; ++i) { for (int j = 0; j < 3; ++j) { newEdges[i][j] = edges[i][j]; } newEdges[i][3] = i; } Arrays.sort(newEdges, new Comparator<int[]>() { public int compare(int[] u, int[] v) { return u[2] - v[2]; } }); UnionFind uf = new UnionFind(n); List<List<Integer>> ans = new ArrayList<List<Integer>>(); for (int i = 0; i < 2; ++i) { ans.add(new ArrayList<Integer>()); } int[] label = new int[m]; for (int i = 0; i < m;) { // 找出所有权值为 w 的边,下标范围为 [i, j) int w = newEdges[i][2]; int j = i; while (j < m && newEdges[j][2] == newEdges[i][2]) { ++j; } // 存储每个连通分量在图 G 中的编号 Map<Integer, Integer> compToId = new HashMap<Integer, Integer>(); // 图 G 的节点数 int gn = 0; for (int k = i; k < j; ++k) { int x = uf.findset(newEdges[k][0]); int y = uf.findset(newEdges[k][1]); if (x != y) { if (!compToId.containsKey(x)) { compToId.put(x, gn++); } if (!compToId.containsKey(y)) { compToId.put(y, gn++); } } else { // 将自环边标记为 -1 label[newEdges[k][3]] = -1; } } // 图 G 的边 List<Integer>[] gm = new List[gn]; List<Integer>[] gmid = new List[gn]; for (int k = 0; k < gn; ++k) { gm[k] = new ArrayList<Integer>(); gmid[k] = new ArrayList<Integer>(); } for (int k = i; k < j; ++k) { int x = uf.findset(newEdges[k][0]); int y = uf.findset(newEdges[k][1]); if (x != y) { int idx = compToId.get(x), idy = compToId.get(y); gm[idx].add(idy); gmid[idx].add(newEdges[k][3]); gm[idy].add(idx); gmid[idy].add(newEdges[k][3]); } } List<Integer> bridges = new TarjanSCC(gn, gm, gmid).getCuttingEdge(); // 将桥边(关键边)标记为 1 for (int id : bridges) { ans.get(0).add(id); label[id] = 1; } for (int k = i; k < j; ++k) { uf.unite(newEdges[k][0], newEdges[k][1]); } i = j; } // 未标记的边即为非桥边(伪关键边) for (int i = 0; i < m; ++i) { if (label[i] == 0) { ans.get(1).add(i); } } return ans; } } // 并查集模板 class UnionFind { int[] parent; int[] size; int n; // 当前连通分量数目 int setCount; public UnionFind(int n) { this.n = n; this.setCount = n; this.parent = new int[n]; this.size = new int[n]; Arrays.fill(size, 1); for (int i = 0; i < n; ++i) { parent[i] = i; } } public int findset(int x) { return parent[x] == x ? x : (parent[x] = findset(parent[x])); } public boolean unite(int x, int y) { x = findset(x); y = findset(y); if (x == y) { return false; } if (size[x] < size[y]) { int temp = x; x = y; y = temp; } parent[y] = x; size[x] += size[y]; --setCount; return true; } public boolean connected(int x, int y) { x = findset(x); y = findset(y); return x == y; } } class TarjanSCC { List<Integer>[] edges; List<Integer>[] edgesId; int[] low; int[] dfn; List<Integer> ans; int n; int ts; public TarjanSCC(int n, List<Integer>[] edges, List<Integer>[] edgesId) { this.edges = edges; this.edgesId = edgesId; this.low = new int[n]; Arrays.fill(low, -1); this.dfn = new int[n]; Arrays.fill(dfn, -1); this.n = n; this.ts = -1; this.ans = new ArrayList<Integer>(); } public List<Integer> getCuttingEdge() { for (int i = 0; i < n; ++i) { if (dfn[i] == -1) { getCuttingEdge(i, -1); } } return ans; } private void getCuttingEdge(int u, int parentEdgeId) { low[u] = dfn[u] = ++ts; for (int i = 0; i < edges[u].size(); ++i) { int v = edges[u].get(i); int id = edgesId[u].get(i); if (dfn[v] == -1) { getCuttingEdge(v, id); low[u] = Math.min(low[u], low[v]); if (low[v] > dfn[u]) { ans.add(id); } } else if (id != parentEdgeId) { low[u] = Math.min(low[u], dfn[v]); } } } }
cpp 解法, 执行用时: 24 ms, 内存消耗: 12.5 MB, 提交时间: 2023-10-10 15:21:00
// 并查集模板 class UnionFind { public: vector<int> parent; vector<int> size; int n; // 当前连通分量数目 int setCount; public: UnionFind(int _n): n(_n), setCount(_n), parent(_n), size(_n, 1) { iota(parent.begin(), parent.end(), 0); } int findset(int x) { return parent[x] == x ? x : parent[x] = findset(parent[x]); } bool unite(int x, int y) { x = findset(x); y = findset(y); if (x == y) { return false; } if (size[x] < size[y]) { swap(x, y); } parent[y] = x; size[x] += size[y]; --setCount; return true; } bool connected(int x, int y) { x = findset(x); y = findset(y); return x == y; } }; // Tarjan 算法求桥边模版 class TarjanSCC { private: const vector<vector<int>>& edges; const vector<vector<int>>& edgesId; vector<int> low; vector<int> dfn; vector<int> ans; int n; int ts; private: void getCuttingEdge_(int u, int parentEdgeId) { low[u] = dfn[u] = ++ts; for (int i = 0; i < edges[u].size(); ++i) { int v = edges[u][i]; int id = edgesId[u][i]; if (dfn[v] == -1) { getCuttingEdge_(v, id); low[u] = min(low[u], low[v]); if (low[v] > dfn[u]) { ans.push_back(id); } } else if (id != parentEdgeId) { low[u] = min(low[u], dfn[v]); } } } public: TarjanSCC(int n_, const vector<vector<int>>& edges_, const vector<vector<int>>& edgesId_): \ edges(edges_), edgesId(edgesId_), low(n_, -1), dfn(n_, -1), n(n_), ts(-1) {} vector<int> getCuttingEdge() { for (int i = 0; i < n; ++i) { if (dfn[i] == -1) { getCuttingEdge_(i, -1); } } return ans; } }; class Solution { public: vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) { int m = edges.size(); for (int i = 0; i < m; ++i) { edges[i].push_back(i); } sort(edges.begin(), edges.end(), [](const auto& u, const auto& v) { return u[2] < v[2]; }); UnionFind uf(n); vector<vector<int>> ans(2); vector<int> label(m); for (int i = 0; i < m;) { // 找出所有权值为 w 的边,下标范围为 [i, j) int w = edges[i][2]; int j = i; while (j < m && edges[j][2] == edges[i][2]) { ++j; } // 存储每个连通分量在图 G 中的编号 unordered_map<int, int> compToId; // 图 G 的节点数 int gn = 0; for (int k = i; k < j; ++k) { int x = uf.findset(edges[k][0]); int y = uf.findset(edges[k][1]); if (x != y) { if (!compToId.count(x)) { compToId[x] = gn++; } if (!compToId.count(y)) { compToId[y] = gn++; } } else { // 将自环边标记为 -1 label[edges[k][3]] = -1; } } // 图 G 的边 vector<vector<int>> gm(gn), gmid(gn); for (int k = i; k < j; ++k) { int x = uf.findset(edges[k][0]); int y = uf.findset(edges[k][1]); if (x != y) { int idx = compToId[x], idy = compToId[y]; gm[idx].push_back(idy); gmid[idx].push_back(edges[k][3]); gm[idy].push_back(idx); gmid[idy].push_back(edges[k][3]); } } vector<int> bridges = TarjanSCC(gn, gm, gmid).getCuttingEdge(); // 将桥边(关键边)标记为 1 for (int id: bridges) { ans[0].push_back(id); label[id] = 1; } for (int k = i; k < j; ++k) { uf.unite(edges[k][0], edges[k][1]); } i = j; } // 未标记的边即为非桥边(伪关键边) for (int i = 0; i < m; ++i) { if (!label[i]) { ans[1].push_back(i); } } return ans; } };
cpp 解法, 执行用时: 84 ms, 内存消耗: 19.2 MB, 提交时间: 2023-10-10 15:20:48
// 并查集模板 class UnionFind { public: vector<int> parent; vector<int> size; int n; // 当前连通分量数目 int setCount; public: UnionFind(int _n): n(_n), setCount(_n), parent(_n), size(_n, 1) { iota(parent.begin(), parent.end(), 0); } int findset(int x) { return parent[x] == x ? x : parent[x] = findset(parent[x]); } bool unite(int x, int y) { x = findset(x); y = findset(y); if (x == y) { return false; } if (size[x] < size[y]) { swap(x, y); } parent[y] = x; size[x] += size[y]; --setCount; return true; } bool connected(int x, int y) { x = findset(x); y = findset(y); return x == y; } }; class Solution { public: vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) { int m = edges.size(); for (int i = 0; i < m; ++i) { edges[i].push_back(i); } sort(edges.begin(), edges.end(), [](const auto& u, const auto& v) { return u[2] < v[2]; }); // 计算 value UnionFind uf_std(n); int value = 0; for (int i = 0; i < m; ++i) { if (uf_std.unite(edges[i][0], edges[i][1])) { value += edges[i][2]; } } vector<vector<int>> ans(2); for (int i = 0; i < m; ++i) { // 判断是否是关键边 UnionFind uf(n); int v = 0; for (int j = 0; j < m; ++j) { if (i != j && uf.unite(edges[j][0], edges[j][1])) { v += edges[j][2]; } } if (uf.setCount != 1 || (uf.setCount == 1 && v > value)) { ans[0].push_back(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 (int j = 0; j < m; ++j) { if (i != j && uf.unite(edges[j][0], edges[j][1])) { v += edges[j][2]; } } if (v == value) { ans[1].push_back(edges[i][3]); } } return ans; } };