P2853 [USACO06DEC]Cow Picnic S

[USACO06DEC]Cow Picnic S

题目描述

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.

K(1≤K≤100)只奶牛分散在N(1≤N≤1000)个牧场．现在她们要集中起来进餐．牧场之间有M(1≤M≤10000)条有向路连接，而且不存在起点和终点相同的有向路．她们进餐的地点必须是所有奶牛都可到达的地方．那么，有多少这样的牧场呢？

输入格式

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.

输出格式

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

样例 #1

样例输入 #1

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

样例输出 #1

2

提示

The cows can meet in pastures 3 or 4.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define de(x) cout<<x<<" ";
#define sf(x) scanf("%d",&x);
#define Pu puts("");
const int N=1e4+10;
int k,n,m;
vector<int>v[N];
queue<int>ox;
int cnt[N];
// int in1[N],in[N];
//受上一题最长路的影响，误以为需要拓扑排序，而实际上并不是这样
//因为上一题边上有权值，越过某个路径（走短路）一定会导致路径变短
//然而此题只需要遍历所有能到达的点即可，哪怕不按照拓扑排序来
int vis[N];
void bfs(){
    int t,o;
    while(!ox.empty()){
        t=ox.front();ox.pop();
        queue<int>q;
        memset(vis,0,sizeof(vis));

        q.push(t);cnt[t]++;//注意本点也需要加
        // for(int i=1;i<=n;i++){
        //     in[i]=in1[i];
        // }
        while(!q.empty()){
            t=q.front();q.pop();vis[t]=1;
            for(int i=0;i<v[t].size();i++){
                o=v[t][i];
                if(!vis[o]){//
                    vis[o]=1;
                    cnt[o]++;
                    q.push(o);
                }
                /*//这是更改了拓扑排序之后的写法
                1、忘记了把访问过的点给赋值为1，即标记为已经访问
                2、在改正1的基础上，由于某些点人会指向下一个点，但是此点已经被访问过了
                //直接cnt[o]++ 显然多加了，是错误的
                o=v[t][i];
                cnt[o]++;
                if(!vis[o]){
                    q.push(o);
                }
                */
                //in[o]--;
                //if(!in[o]) q.push(o);
            }
        }
    }
}
int main(){
    cin>>k>>n>>m;
    int x,y;
    for(int i=1;i<=k;i++){
        sf(x)
        ox.push(x);
    }
    for(int i=1;i<=m;i++){
        sf(x)sf(y)
        v[x].push_back(y);
        //in1[y]++;
    }
    bfs();
    int ans=0;
    for(int i=1;i<=m;i++){
        if(cnt[i]==k) ans++;
    }
    printf("%d\n",ans);
    return 0;
}