[C++知识库]Tarjan算法例题及模板

#include<bits/stdc++.h>
using namespace std;
const int N = 10010 , M = 50010;
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;  // 时间戳
int stk[N], top;
bool in_stk[N];
int id[N], scc_cnt;  // 每个点所属分量编号
int dout[N], Size[N];   // 出度数组 和 统计连通分量里面点的个数的数组

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

void tarjan(int u)
{
    dfn[u] = low[u] = ++ timestamp;
    stk[ ++ top] = u, in_stk[u] = true;
    
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (!dfn[j])
        {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        }
        else if (in_stk[j])
            low[u] = min(low[u], dfn[j]);
    }
    
    if (dfn[u] == low[u])
    {
        ++ scc_cnt;
        int y;
        do {
            y = stk[top -- ];
            in_stk[y] = false;
            id[y] = scc_cnt;
            Size[scc_cnt] ++;
        } while (y != u);
    }
}


int main()
{
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);
    while (m -- ){
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);
    }
    
    for(int i = 1; i <= n; i ++)
        if(!dfn[i])
            tarjan(i);
    
    for (int i = 1; i <= n; i ++ )
        for(int j = h[i]; ~j ; j = ne[j])
        {
            int k = e[j];
            int a = id[i] , b = id[k];
            if (a != b) dout[a] ++;
        }
    
    int zeros = 0, sum = 0;
    for (int i = 1; i <= scc_cnt; i ++ )
        if(!dout[i])
        {
            zeros ++;
            sum += Size[i];
            if(zeros > 1)
            {
                sum = 0;
                break;
            }
        }
        
    printf("%d\n", sum);
    return 0;
}