有向图的强连通分量

两次dfs模板：

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define M(a, b) memset(a, b, sizeof(a))
const int N = 1000 + 5;
int vis[N], sccno[N], scc_cnt;
vector<int> G[N], G2[N];
vector<int> S;

void dfs1(int u) {
    if (vis[u]) return;
    vis[u] = 1;
    for (int i = 0; i < G[u].size(); ++i) dfs1(G[u][i]);
    S.push_back(u);
}

void dfs2(int u) {
    if (sccno[u]) return;
    sccno[u] = scc_cnt;
    for (int i = 0; i < G[u].size(); ++i) dfs2(G[u][i]);
}

void find_scc(int n) {
    M(vis, 0); M(sccno, 0);
    S.clear(); scc_cnt = 0;
    for (int i = 0; i < n; ++i) dfs1(i);
    for (int i = n-1; i >= 0; --i) 
        if (!sccno[i]) {++scc_cnt; dfs2(i);}
}

int main() {
    int n, m;
    while (cin >> n >> m) {
        int u, v;
        for (int i = 0; i < m; ++i) {
            cin >> u >> v;
            G[u].push_back(v);
            G2[v].push_back(u);
        }
        find_scc(n);        
    }

    return 0;
}

Tarjan算法模板：

#include<bits/stdc++.h>
using namespace std;
#define M(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f
const int N = 1000 + 5;
int pre[N], sccno[N], dfs_clock, scc_cnt;
vector<int> G[N];
stack<int> S;

int dfs(int u) {
    int lowu = pre[u] = ++dfs_clock;
    S.push(u);
    for (int i = 0; i < G[u].size(); ++i) {
        int v = G[u][i];
        if (!pre[v]) {
            int lowv = dfs(v);
            lowu = min(lowu, lowv);
        }
        else if (!sccno[v]) { //加这个判定条件去掉已经包含在其他强连通分量中的反向边
            lowu = min(lowu, pre[v]);
        }
    }
    if (lowu == pre[u]) {
        ++scc_cnt;
        while (true) {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if (x == u) break;
        }
    }
    return lowu;
}

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

int main() {
    int n, m;
    while (cin >> n >> m) {
        int u, v;
        for (int i = 0; i < m; ++i) {
            cin >> u >> v;
            G[u].push_back(v);
        }
        find_scc(n);        
    }

    return 0;
}