poj 2186(强连通分量)

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#integer #numbers #output #pair #input #up

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本文介绍了一种通过图论中的强连通分量算法解决牛群流行度计算问题的方法。在一个给定数量的牛群中，通过每头牛对其他牛的流行度评价构建有向图，并利用Tarjan算法进行强连通分量分析，最终计算出被所有牛认为流行的牛的数量。

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14393		Accepted: 5718

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： http://poj.org/problem?id=2186

分析：这题首先想到的就是求强连通缩点，然后计算出度为0的点的个数，如果为1，即答案所求，否则无解，输出0；

代码：

#include<cstdio>
#define min(a,b) a<b?a:b
using namespace std;
const int mm=50005;
const int mn=10001;
int s[mm],t[mm],p[mm];
int h[mn],id[mn],q[mn],dfn[mn],low[mn],num[mn];
int i,j,k,n,m,tsp,qe,cnt;
void dfs(int u)
{
    int i,v;
    dfn[u]=low[q[qe++]=u]=++tsp;
    for(i=h[u];i>=0;i=p[i])
        if(!dfn[v=t[i]])
            dfs(v),low[u]=min(low[u],low[v]);
        else if(id[v]<0)low[u]=min(low[u],dfn[v]);
    if(low[u]==dfn[u])
    {
        num[id[u]=++cnt]=1;
        while((v=q[--qe])!=u)++num[id[v]=cnt];
    }
}
void tarjan()
{
    int i;
    for(tsp=qe=cnt=i=0;i<=n;++i)id[i]=-1,dfn[i]=0;
    for(i=1;i<=n;++i)
        if(!dfn[i])dfs(i);
}
int main()
{
    while(scanf("%d%d",&n,&m)!=-1)
    {
        for(i=0;i<=n;++i)h[i]=-1;
        for(k=0;k<m;++k)
        {
            scanf("%d%d",&i,&j);
            s[k]=i,t[k]=j,p[k]=h[i],h[i]=k;
        }
        tarjan();
        for(i=0;i<=cnt;++i)q[i]=0;
        for(i=0;i<m;++i)
            if((j=id[s[i]])!=id[t[i]])++q[j];
        for(tsp=0,i=1;i<=cnt;++i)
            if(!q[i])++tsp,j=i;
        if(tsp>1)printf("0\n");
        else printf("%d\n",num[j]);
    }
    return 0;
}