Problem 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

題意：k 頭牛，n 個牧場，m 條單向的路，求所有牛能到達(dá)的牧場數(shù)

思路：對每頭牛進(jìn)行 dfs 搜索，最后統(tǒng)計每個農(nóng)場到達(dá)的牛數(shù)。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<set> #include<map> #include<stack> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 10001 #define MOD 123 #define E 1e-6 using namespace std; int k,n,m; int cow[N]; int sum[N]; int vis[N]; vector<int> g[N]; void dfs(int i) {vis[i]=1;//標(biāo)記為已到達(dá)int len=g[i].size();//從當(dāng)前牧場可到達(dá)其他牧場的個數(shù)for(int j=0;j<len;j++)//枚舉所有可到達(dá)的牧場{int next=g[i][j];if(!vis[next])//如果下一個牧場未到達(dá)，繼續(xù)向下搜索dfs(next);} } int main() {scanf("%d%d%d",&k,&n,&m);for(int i=1;i<=k;i++)scanf("%d",&cow[i]);while(m--)//vector鄰接表建圖{int x,y;scanf("%d%d",&x,&y);g[x].push_back(y);}for(int i=1;i<=k;i++){dfs(cow[i]);//dfs每一頭牛for(int j=1;j<=n;j++)//統(tǒng)計每一個牧場的到達(dá)數(shù){sum[j]+=vis[j];vis[j]=0;}}int cnt=0;for(int i=1;i<=n;i++)//枚舉所有牧場if(sum[i]==k)//如果該牧場所有牛均能到達(dá)cnt++;printf("%d\n",cnt);return 0; }

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