[HAOI2006]受欢迎的牛（tarjan缩点）

洛谷传送门

直接tarjan求scc，然后统计出度为0的缩点，如果多余1个就输出0，只有一个就输出这个缩点里的点。

——代码

#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;

int n, m, cnt, idx, ans, k;
int next[100001], to[100001], head[10001], low[10001], dfn[10001], belong[10001], c[10001];
bool ins[10001];
stack <int> s;

inline void add(int x, int y)
{
    to[cnt] = y;
    next[cnt] = head[x];
    head[x] = cnt++;
}

inline void tarjan(int u)
{
    low[u] = dfn[u] = ++idx;
    s.push(u);
    ins[u] = 1;
    int i, v;
    for(i = head[u]; i != -1; i = next[i])
    {
        v = to[i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v]) low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u])
    {
        cnt++;
        do
        {
            v = s.top();
            s.pop();
            ins[v] = 0;
            belong[v] = cnt;
        }while(u != v);
    }
}

int main()
{
    int i, j, x, y, u, v;
    memset(head, -1, sizeof(head));
    scanf("%d %d", &n, &m);
    for(i = 1; i <= m; i++)
    {
        scanf("%d %d", &x, &y);
        add(x, y);
    }
    cnt = 0;
    for(i = 1; i <= n; i++)
     if(!dfn[i])
      tarjan(i);
    for(u = 1; u <= n; u++)
     for(i = head[u]; i != -1; i = next[i])
     {
         v = to[i];
         if(belong[u] != belong[v]) c[belong[u]]++;
     }
    for(i = 1; i <= cnt; i++)
     if(!c[i])
     {
         ans++;
         k = i;
     }
    if(ans > 1)    printf("0");
    else
    {
        ans = 0;
        for(i = 1; i <= n; i++)
         if(belong[i] == k)
          ans++;
        printf("%d", ans);
    }
    return 0;
}