poj2816

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29799		Accepted: 12090

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

tarjan模板

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<stack>
using namespace std;
#define N 10010
vector<int>grap[N];//稀疏图，用邻接表表示图
stack<int>s;//栈 
int low[N];//low[u] 为u或u的子树能够追溯到的最早的栈中节点的次序编号  
int dfn[N];//dfn[u] 为u搜索的次序编号(时间戳)  
int mark[N];//标记是否在栈中  
int id[N];//id[i] = j 表示原图的点i缩点后为点j  
int pd;//顶点的前序编号
int sd;//记录总共将图缩成多少个点 
int sum[N];//记录sd编号的有几个点缩成 
void tarjan(int v){
    low[v]=dfn[v]=++pd;
    s.push(v); 
    mark[v]=1;
    for(int i=0;i<grap[v].size();i++){
        int w=grap[v][i];
        if(!dfn[w]){
            tarjan(w);
            low[v]=min(low[v],low[w]);//v或v的子树能够追溯到的最早的栈中节点的次序编号 
        }
        else if(mark[w]){//(v,w)为后向边
            low[v]=min(low[v],dfn[w]);
        }    
    }
    int u;
    if(low[v]==dfn[v]){//满足强连通分支条件,进行缩点 
        sd++;
        do{
            u=s.top();
            s.pop();
            id[u]=sd;//缩点
            sum[sd]++;
            mark[u]=0;//出栈解除标记

        }while(u!=v);
    }
}
int main(){
    int n,m,a,b;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        grap[a].push_back(b);
    }
    for(int i=1;i<=n;i++){
        if(!dfn[i]) tarjan(i);
    }
    //if(sd==1) {printf("0
");return 0;}//如果图已经为强连通图,over 
    int in[N]={0},out[N]={0};
    for(int i=1;i<=n;i++){//求缩点后，各个顶点的出度和入度 
        for(int j=0;j<grap[i].size();j++){
            int k=grap[i][j];
            if(id[i]!=id[k]){
                in[id[k]]++;
                out[id[i]]++;
            }
        }
    }
    int ans=0,p=0;
    for(int i=1;i<=sd;i++){
        if(!out[i]){
            ans++;p=i;
        }
    }
    printf("%d
",ans==1?sum[p]:0);
    return 0;
}