POJ 3256 Cow Picnic

Cow Picnic

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4928		Accepted: 2019

Description

The cows are having a picnic! Each of Farmer John's K (1 ≤ K ≤ 100) cows is grazing in one of N (1 ≤ N ≤ 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 ≤ M ≤ 10,000) one-way paths (no path connects a pasture to itself).

The cows want to gather in the same pasture for their picnic, but (because of the one-way paths) some cows may only be able to get to some pastures. Help the cows out by figuring out how many pastures are reachable by all cows, and hence are possible picnic locations.

Input

Line 1: Three space-separated integers, respectively: K, N, and M 
Lines 2..K+1: Line i+1 contains a single integer (1..N) which is the number of the pasture in which cow i is grazing. 
Lines K+2..M+K+1: Each line contains two space-separated integers, respectively A and B (both 1..N and A != B), representing a one-way path from pasture A to pasture B.

Output

Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.

Sample Input

2 4 4
2
3
1 2
1 4
2 3
3 4

Sample Output

2

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;
}