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\n", 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
***/