关键词：
题目连接：
https://www.luogu.com.cn/problem/P2853
题意：
有n个奶牛在不同牧场，牧场之间有m条路。
求所以奶牛都能共同到达的牧场的数量
思路：
dfs每一个奶牛可以到的牧场。
开一个d[N]数组记录d[i] i这个牧场能来的奶牛的个数，如果d[i]==n，就符合题意。
AC代码：
#include<bits/stdc++.h>
using namespace std;
const int N=101000;
vector<int> p[N];
int n,m,k;
int ans;
int a[N];
int d[N];
bool st[N];
void dfs(int u)
// if(p[u].size()==0) return ;
// printf("u===%d \\n",u);
d[u]++;
st[u]=true;
for(int i=0;i<p[u].size();i++)
int t=p[u][i];
// printf("u===%d t===%d \\n",u,t);
if(!st[t])
// d[t]++;
dfs(t);
int main()
cin>>n>>m>>k;
for(int i=1;i<=n;i++)
cin>>a[i];
while(k--)
int x,y;
cin>>x>>y;
p[x].push_back(y);
// p[y].push_back(x);
for(int i=1;i<=n;i++)
memset(st,0,sizeof(st));
dfs(a[i]);
for(int i=1;i<=m;i++)
// printf("d[%d]==%d \\n",i,d[i]);
if(d[i]==n)
ans++;
cout<<ans<<endl;
```
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