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.

第1行輸入K，N，M.接下來K行，每行一個整數表示一只奶牛所在的牧場編號．接下來M行，每行兩個整數，表示一條有向路的起點和終點

Output

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

所有奶牛都可到達的牧場個數

Sample Input

Sample Output

HINT

Source

Silver

Analysis

DFS模擬奶牛走路就行了

Code

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #define maxn 100000
 5 using namespace std;
 6 
 7 struct edge{
 8     int from,v;
 9 }e[maxn];
10 
11 int mark[maxn],cow[maxn],k,n,m;
12 bool book[maxn];
13 
14 int tot,first[maxn];
15 void insert(int u,int v){tot++;e[tot].from = first[u],e[tot].v = v,first[u] = tot;}
16 
17 void dfs(int now){
18     mark[now]++;
19     for(int i = first[now];i;i = e[i].from){
20         int v = e[i].v;
21         if(!book[v]){
22             book[v] = true;
23             dfs(v);
24         }
25     }
26 }
27 
28 int main(){
29     scanf("%d%d%d",&k,&n,&m);
30     
31     for(int i = 1;i <= k;i++)
32         scanf("%d",&cow[i]);
33     
34     for(int i = 1;i <= m;i++){
35         int a,b;
36         scanf("%d%d",&a,&b);
37         insert(a,b);
38     }
39     
40     for(int i = 1;i <= k;i++){
41         memset(book,false,sizeof(book));
42         book[cow[i]] = true;
43         dfs(cow[i]);
44     }
45     
46     int ans = 0;
47     for(int i = 1;i <= n;i++) if(mark[i] == k) ans++;
48     cout << ans;
49     
50     return 0;
51 }

[BZOJ] 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐