75 lines
2.3 KiB
C++
75 lines
2.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 5e5 + 10, M = 4e6 + 10;
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//链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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//点双需要的变量
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int dfn[N], low[N], stk[N], timestamp, top;
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vector<int> bcc[N];
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int bcnt;
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//点双模板1 【推荐】
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void tarjan(int u, int fa) {
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low[u] = dfn[u] = ++timestamp;
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stk[++top] = u;
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int son = 0; //子节点个数
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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// if (j == fa) continue; //这句看似无用，dfn[j]可以包含，但事实证明，有些题会在后面加入其它代码造成执行逻辑错误，背模板时，加上这句保准！
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if (!dfn[j]) {
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son++; //找到一个子节点
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tarjan(j, u);
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low[u] = min(low[u], low[j]);
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if (low[j] >= dfn[u]) {
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int x;
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bcnt++;
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do {
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x = stk[top--];
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bcc[bcnt].push_back(x);
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} while (x != j); //将子树出栈
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bcc[bcnt].push_back(u); //把割点/树根也丢到点双里
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}
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} else
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low[u] = min(low[u], dfn[j]);
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}
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//特判独立点
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if (fa == -1 && son == 0) bcc[++bcnt].push_back(u);
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}
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int n, m;
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int main() {
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//文件输入输出
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#ifndef ONLINE_JUDGE
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freopen("P8435.in", "r", stdin);
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#endif
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//清空链式前向星
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m);
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while (m--) {
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int a, b;
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scanf("%d %d", &a, &b);
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if (a != b) add(a, b), add(b, a); //此题目需要判断一下自环，否则1和11会挂
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}
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//给定一个n个点m条边的无向图，求该图中的所有点双连通分量(v-bcc)
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for (int i = 1; i <= n; i++) //每个点做为根结点进行枚举
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if (!dfn[i]) tarjan(i, -1);
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printf("%d\n", bcnt);
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for (int i = 1; i <= bcnt; i++) { //枚举每个双连通分量
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printf("%d ", bcc[i].size());
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for (int j = 0; j < bcc[i].size(); j++) //输出双连通分量中的点有哪些
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printf("%d ", bcc[i][j]);
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puts("");
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}
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return 0;
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} |