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#include <bits/stdc++.h>
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using namespace std;
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//此题是P8436的弱化版本，弱化了记录每个边双中点有哪些:vector<int> belong[N];
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const int N = 500010, M = 2000010 * 2;
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int n, m;
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int dfn[N], low[N], ts, stk[N], top;
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vector<int> belong[N]; //边双中有哪些原始点
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int id[N], dcc_cnt; //原始点x属于哪个边双连通分量，dcc_cnt指边双连通分量个数
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int is_bridge[M]; //记录哪些边是割边
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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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//边双连通分量
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void tarjan(int u, int fa) {
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dfn[u] = low[u] = ++ts;
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stk[++top] = u;
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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 (!dfn[j]) {
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tarjan(j, i);
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low[u] = min(low[u], low[j]);
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if (dfn[u] < low[j]) is_bridge[i] = is_bridge[i ^ 1] = true; //记录割边
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} else if (i != (fa ^ 1))
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low[u] = min(low[u], dfn[j]);
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}
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if (dfn[u] == low[u]) {
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++dcc_cnt; //边双数量+1
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int x;
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do {
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x = stk[top--];
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id[x] = dcc_cnt; // 记录点与边双关系
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belong[dcc_cnt].push_back(x); // 记录边双中有哪些点
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} while (x != u);
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}
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}
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int main() {
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scanf("%d %d", &n, &m);
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memset(h, -1, sizeof h);
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for (int i = 0; i < m; i++) {
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int a, b;
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scanf("%d %d", &a, &b);
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add(a, b), add(b, a);
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}
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i, -1);
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//输出边双个数
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printf("%d\n", dcc_cnt);
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// for (int i = 1; i <= dcc_cnt; i++) {
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// //此边双中点的数量
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// printf("%d ", belong[i].size());
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// //此边双中都有哪些点
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// for (int j = 0; j < belong[i].size(); j++)
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// printf("%d ", belong[i][j]);
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// puts("");
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// }
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return 0;
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} |
