洛谷 P2853 [USACO06DEC]牛的野餐Cow Picnic

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最新推荐文章于 2025-07-03 11:02:04 发布

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这篇博客介绍了USACO竞赛中的一道题目，涉及奶牛在有向图中寻找所有奶牛都能到达的牧场进行野餐的问题。通过输入输出样例和说明，解释了如何计算可达的牧场数量，提出了反向建图和DFS遍历的方法来解决这个问题。

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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？

输入输出格式

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.

输入输出样例

输入样例#1：

输出样例#1：

说明

The cows can meet in pastures 3 or 4.

反向建图，对于每一个点dfs统计可以到达它的牛的个数。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=1005;
const int M=10005;
int k,n,m,cnt,ans,hd[N],a[N];
bool b[N];
struct edge
{
	int to,nxt;
}v[M];
void addedge(int x,int y)
{
	++cnt;
	v[cnt].to=y;
	v[cnt].nxt=hd[x];
	hd[x]=cnt;
}
void dfs(int u)
{
	b[u]=1;
	for(int i=hd[u];i;i=v[i].nxt)
		if(!b[v[i].to])
			dfs(v[i].to);
}
int main()
{
	scanf("%d%d%d",&k,&n,&m);
	for(int i=1;i<=k;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		addedge(y,x);
	}
	for(int i=1;i<=n;i++)
	{
		int sum=0;
		memset(b,0,sizeof(b));
		dfs(i);
		for(int j=1;j<=k;j++)
			if(b[a[j]])
				++sum;
		if(sum==k)
			++ans;
	}
	printf("%d\n",ans);
	return 0;
}