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