思路：DFS全图，记录每个牧场可以到达的牛的数量，若pa[v] == K，则所有牛可以到达。

#include
     
      #include
      
       #include
       
        #define MAX 10005using namespace std;typedef struct{    int to, next;}Node;Node edge[MAX];int head[1005], vis[1005], pa[MAX], in[MAX];void dfs(int s){    for(int i = head[s];i != -1;i = edge[i].next){        int v = edge[i].to;        if(!vis[v]){            pa[v] ++;            vis[v] = 1;            dfs(v);        }    }}int main(){    int K, N, M, u, v;    /* freopen("in.c", "r", stdin); */    while(~scanf("%d%d%d", &K, &N, &M)){        memset(head, -1, sizeof(head));        memset(pa, 0, sizeof(pa));        for(int i = 1;i <= K;i ++){            scanf("%d", &in[i]);            pa[in[i]] ++;        }        for(int i = 1;i <= M;i ++){            scanf("%d%d", &u, &v);            edge[i].to = v;            edge[i].next = head[u];            head[u] = i;        }        for(int i = 1;i <= K;i ++){            memset(vis, 0, sizeof(vis));            vis[in[i]] = 1;            dfs(in[i]);        }        int ans = 0;        for(int i = 1;i <= N;i ++)            if(pa[i] == K) ans ++;        cout << ans << endl;    }    return 0;}