64 lines
1.5 KiB
C++
64 lines
1.5 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 100010, M = 2000010; // 因为要建新图，两倍的边
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int n, m, mod; // 点数、边数、取模的数
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int f[N], g[N];
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int h[N], hs[N], e[M], ne[M], idx; // h: 原图；hs: 新图
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void add(int a, int b) {
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e[idx] = b, ne[idx] = h[a], h[a] = idx++;
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}
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// tarjan求强连通分量
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int dfn[N], low[N], ts, in_stk[N], stk[N], top;
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int id[N], scc_cnt, sz[N];
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void tarjan(int u) {
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dfn[u] = low[u] = ++ts;
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stk[++top] = u;
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in_stk[u] = 1;
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (!dfn[v]) {
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tarjan(v);
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low[u] = min(low[u], low[v]);
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} else if (in_stk[v])
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low[u] = min(low[u], low[v]);
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}
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if (dfn[u] == low[u]) {
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++scc_cnt;
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int x;
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do {
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x = stk[top--];
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in_stk[x] = 0;
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id[x] = scc_cnt;
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sz[scc_cnt]++;
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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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#ifndef ONLINE_JUDGE
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freopen("1175_Preapre.in", "r", stdin);
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#endif
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memset(h, -1, sizeof h); // 原图
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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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add(a, b);
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}
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// (1) tarjan算法求强连通分量
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i);
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for (int i = 1; i <= n; i++)
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cout << "id[" << i << "]=" << id[i] << " ";
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cout << endl;
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cout << "scc_cnt=" << scc_cnt << endl;
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return 0;
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} |