强连通分量（LRJ训练指南）

#include <iostream>
#include <queue>
#include <string>
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
const int maxn = 10000;
int gn, gm;

vector<int> gg[maxn];
int pre[maxn], low[maxn], sccno[maxn], dfs_clock, scc_cnt;

stack<int> gs;

void dfs(int u) {
    int i;
    pre[u] = low[u] = ++dfs_clock;
    gs.push(u);
    for(i = 0; i < (int)gg[u].size(); i++) {
        int v = gg[u][i];
        if(!pre[v]) {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if(!sccno[v]) {
            low[u] = min(low[u], low[v]);//min(low[u], low[v]);
        }
    }
    if(low[u] == pre[u]) {
        scc_cnt++;
        for(;;) {
            int x = gs.top();
            gs.pop();
            sccno[x] = scc_cnt;
            if(x == u) break;
        }
    }
}

void find_scc(int n) {
    dfs_clock = scc_cnt = 0;
    memset(sccno, 0, sizeof(sccno));
    memset(pre, 0, sizeof(pre));
    for(int i = 1; i <= n; i++) {
        if(!pre[i]) dfs(i);
    }
}

void print() {
    for(int i = 1; i <= gn; i++) {
        printf("i = %d -> %d
", i, sccno[i]);
    }
}

int main()
{
    int from, to;
    while(scanf("%d%d", &gn, &gm) ==2) {
        for(int i = 0; i < gm; i++) {
            scanf("%d%d", &from, &to);
            gg[from].push_back(to);
        }
        find_scc(gn);
        print();
    }
    return 0;
}

/**
11 16
1 2
2 3
2 5
3 4
4 3
4 7
5 2
5 4
5 6
6 7
6 10
7 8
8 9
9 10
10 11
11 6
***/