1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 514 Solved: 316
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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行输入K，N，M.接下来K行，每行一个整数表示一只奶牛所在的牧场编号．接下来M行，每行两个整数，表示一条有向路的起点和终点
Output
* Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.
所有奶牛都可到达的牧场个数
Sample Input
2
3
1 2
1 4
2 3
3 4
INPUT DETAILS:
4<--3
^ ^
| |
| |
1-->2
The pastures are laid out as shown above, with cows in pastures 2 and 3.
Sample Output
牧场3，4是这样的牧场．
HINT
Source
Silver
题解：从每个奶牛dfs，记录是否可以到达。枚举n个牧场。
五个字：暴力大法好。
<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define ll long long
using namespace std;
int read()
{int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
struct Node{int to,next;
}e[10005];
int head[1005],tot=0,K,N,M,ans,a[105];
bool v[105][1005];
void add(int u,int v)
{e[++tot]=(Node){v,head[u]};head[u]=tot;
}
void dfs(int num,int x)
{v[num][x]=1;for(int i=head[x];i;i=e[i].next){if(!v[num][e[i].to])dfs(num,e[i].to);}
}
int main()
{K=read();N=read();M=read();for(int i=1;i<=K;i++)a[i]=read();int x,y;for(int i=1;i<=M;i++){x=read();y=read();add(x,y);}for(int i=1;i<=K;i++)dfs(i,a[i]);for(int i=1;i<=N;i++){bool ok=1;for(int j=1;j<=K;j++)if(!v[j][i]){ok=false;break;}if(ok==1)ans++;}cout<<ans<<endl;
}</span>
