POJ 3256

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<iostream>
#include<cstdio>
#include<cstring>
#define MAX 10005
using 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;
}