【POJ 2186】Popular Cows

weixin_30621919

于 2019-07-09 00:08:00 发布

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原文链接：http://www.cnblogs.com/wuhu-JJJ/p/11154906.html

博客围绕牛受欢迎问题展开，题目是有N头牛，给出M对整数(A,B)表示牛A认为牛B受欢迎。题解先用tarjan求出每个强连通分量，再缩点，统计每个点出度，若有且只有1个出度为0的点，输出该点包含节点数，否则输出0。

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Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 42555		Accepted: 17301

Description

Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.

Input

* Line 1: Two space-separated integers, N and M

* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow.

Sample Input

Sample Output

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

题目大意：每一头牛的愿望就是变成一头最受欢迎的牛。现在有N头牛，给你M对整数(A,B)，表示牛A认为牛B受欢迎。这

种关系是具有传递性的，如果A认为B受欢迎，B认为C受欢迎，那么牛A也认为牛C受欢迎。你的任务是求出有多少头

牛被所有的牛认为是受欢迎的。

题解：还是觉得有点难啊……别人的题解……拿来借鉴看看(真的不是很会写，哎，只会模板)

先用tarjan求出每个强连通分量，再缩点，统计每个点的出度，如果有且只有1个出度为0的点，就输出这个点包含的节点数，否则输出0.

#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
#include<cmath>
using namespace std;
vector<int>g[10010];
int n,m,x,y,i,j,v,c[10010],l=0,low[10010],dfn[10010],f[10010],cnt=0,out0[10010],sum[10010],time_clock=0;
void tarjan(int u){
    low[u]=dfn[u]=++time_clock;
    c[++l]=u;
    for(int i=0;i<g[u].size();++i){
        v=g[u][i];
        if(!dfn[v]){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }else if(!f[v])low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u]){
        int len=l;
        cnt++;
        while(c[l]!=u)f[c[l--]]=cnt;
        f[c[l--]]=cnt;
        sum[cnt]=len-l;
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(i=1;i<=m;++i){
        scanf("%d%d",&x,&y);
        g[x].push_back(y);
    }
    memset(dfn,0,sizeof(dfn));
    for(i=1;i<=n;++i)if(!dfn[i])tarjan(i);
    for(i=1;i<=n;++i)
    for(j=0;j<g[i].size();++j){
        v=g[i][j];
        if(f[i]!=f[v])out0[f[i]]++;
    }
    x=0;
    for(i=1;i<=cnt;++i)
    if(!out0[i]){
        if(x>0){
            printf("0");
            return 0;
        }
        x=sum[i];
    }
    printf("%d",x);
    return 0;
}