求強連通分支 tarjan算法

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
const int maxn = 10;
int tree[maxn];
int dfn[maxn];
vector<int>g[maxn];
int n, m;
int temp[maxn];
int top;
int sta[maxn];
void init() {
   for (int i = 1; i <= n; i ++) {
      g[i].clear();
      temp[i] = 0;
   }
}

void tarbfs(int k, int lay, int& scc_num) {
     temp[k] = 1;
     tree[k] = lay;
     dfn[k] = lay;
     sta[++top] = k;
     for (int i = 0; i < (int)g[k].size(); i ++) {
        if (temp[g[k][i]] == 0) {
            tarbfs(g[k][i], ++lay, scc_num);
        }
        if (temp[g[k][i]] == 1) tree[k] = min(tree[k], tree[g[k][i]]);
     }
     if (dfn[k] == tree[k]) {
        ++scc_num;
        do {
            tree[sta[top]] = k;  //極大強連通分支縮點
            temp[sta[top]] = 2;
        }while (sta[top--] != k);
     }
}
int tarjan() {
    int lay = 1;
    int scc_num = 0;
    top = 0;
    for (int i = 1; i <= n; i ++) {
        if (temp[i] == 0) {
            tarbfs(i, lay, scc_num);
        }
    }
return scc_num;
}
int main() {
    while (cin >> n >> m) {
        init();
        int x, y;
        for (int i = 0; i < m; i ++) {
            scanf("%d%d", &x, &y);
            g[x].push_back(y);
        }
        cout << tarjan() <<endl;
        for (int i = 1; i <= n; i ++) {
            cout << i<<"---->"<<tree[i]<<endl;
        }
    }
return 0;
}