POJ 2186 Popular Cows

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21685		Accepted: 8852

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

3 3
1 2
2 1
2 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity. 

Source

USACO 2003 Fall

初学强连通分量第一题

假设有两头牛A和B都被其他所有牛认为是红人，那么显然A、B也互相认为是红人，即存在一个包含A、B两个顶点的圈，即A和B同属于同一个强连通分量。由此我们可知，如果一头牛被其它所有牛认为是红人，那么它所在的强连通分量内的所有牛都被其他所有牛认为是红人。这样我们把图进行强连通分解后，至多有一个强连通分量满足题目的条件。

我们知道，把一个图进行强连通分量分解后，可以得到各个强连通分量拓扑排序后的顺序，而在此题中，唯一可能成为解的只有拓扑序最后的一个强连通分量。因此我们在对图进行强连通分量分解后，只要检查最后一个强连通分量中的一个顶点是否从所有顶点到达就好了。

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#define MAX_V 10050

using namespace std;

int V,E;
vector<int> G[MAX_V];
vector<int> rG[MAX_V];
vector<int> vs;
bool used[MAX_V];
int cmp[MAX_V];

void add_edge(int _from,int _to)
{
    G[_from].push_back(_to);
    rG[_to].push_back(_from);
}

void dfs(int v)
{
    used[v]=true;
    for(int i=0;i<G[v].size();i++)
        if(!used[G[v][i]])
            dfs(G[v][i]);
    vs.push_back(v);
}

void rdfs(int v,int k)
{
    used[v]=true;
    cmp[v]=k;
    for(int i=0;i<rG[v].size();i++)
        if(!used[rG[v][i]])
            rdfs(rG[v][i],k);
}

int scc()
{
    memset(used,false,sizeof(used));
    vs.clear();
    for(int v=0;v<V;v++)
        if(!used[v])
            dfs(v);

    memset(used,false,sizeof(used));
    int k=0;
    for(int i=vs.size()-1;i>=0;i--)
        if(!used[vs[i]])
        {
            rdfs(vs[i],k);
            k++;
        }

    return k;
}

int main()
{
    while(scanf("%d %d",&V,&E)==2)
    {
        for(int i=0;i<V;i++)
        {
            G[i].clear();
            rG[i].clear();
        }

        int a,b;
        for(int i=1;i<=E;i++)
        {
            scanf("%d %d",&a,&b);
            add_edge(a-1,b-1);
        }

        int k=scc()-1;
        int u,ans=0;

        for(int i=0;i<V;i++)
            if(cmp[i]==k)
            {
                u=i;
                ans++;
            }

        //再反向图上再进行一次DFS，判断最后一个强连通分量中的点是否是从所有点都可达的
        memset(used,false,sizeof(used));
        if(ans)
        {
            rdfs(u,0);
            for(int i=0;i<V;i++)
                if(!used[i])
                {
                    ans=0;
                    break;
                }
        }

        printf("%d
",ans);
    }

    return 0;
}