poj 2186 Popular Cows（強連通分量）

題目：​​http://poj.org/problem?id=2186​​

Popular Cows

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 27673	Accepted: 11139

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

分析：

題目大意：如果A牛崇拜B牛，B牛崇拜C牛，則A牛也崇拜C牛。尋找有多少頭牛它被所有的其他的牛崇拜。

關于強連通圖：如果有向圖G的任何兩頂點都互相可達，則稱圖G是強連通圖，如果有向圖G存在兩頂點u和v，使得u不能到達v或者v不能到達u，則稱圖G是非強連通圖。

在有向無環圖中有這樣的點很特殊：outdegree=0 和 indegree=0的點。嘗試将強連通分量壓縮，通過縮點把雜亂的有向圖變成一幅有向無環圖，出度為0的點個數設為k，如果k=1那麼就表明整個圖是連通的，結果就是壓縮點的個數，k>1則表明圖不是連通的，結果是0。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int M=5e4+10,N=1e4+10;
int head[N],ed;
struct node{
    int to,next;
}edge[M];
int sta[N],vis[N],low[N],dfn[N],out[N],top;
void init(){
    ed=0;
    memset(head,-1,sizeof(head));
    memset(edge,0,sizeof(edge));
    memset(out,0,sizeof(out));
    memset(sta,0,sizeof(sta));
}
void addedge(int a,int b){
    edge[ed].to=b;
    edge[ed].next=head[a];
    head[a]=ed++;
}
void tarbfs(int k,int cnt,int &num){
    vis[k]=1;
    low[k]=cnt;
    dfn[k]=cnt;
    sta[top++]=k;
    for(int i=head[k];i>-1;i=edge[i].next){
        if(vis[edge[i].to]==0) tarbfs(edge[i].to,++cnt,num);
        if(vis[edge[i].to]==1) low[k]=min(low[k],low[edge[i].to]);
    }
    if(dfn[k]==low[k]){
        ++num;
        while(top>0&&sta[top]!=k){
            top--;
            low[sta[top]]=num;
            vis[sta[top]]=2;
        }
    }
}
int tarjan(int n){
    int num=0,cnt=1;
    top=0;
    memset(vis,0,sizeof(vis));
    memset(low,0,sizeof(low));
    for(int i=1;i<=n;i++){
        if(vis[i]==0) tarbfs(i,cnt,num);
    }
    return num;
}

int main()
{
    //freopen("cin.txt","r",stdin);
    int n,m;
    while(cin>>n>>m){
        int a,b;
        init();
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            addedge(a,b);
        }
        int num=tarjan(n); 
        for(int i=1;i<=n;i++){
            for(int j=head[i];j>-1;j=edge[j].next){
                if(low[i]!=low[edge[j].to]) out[low[i]]++;
            }
        }
        int sum=0,x;
        for(int i=1;i<=num;i++){
            if(out[i]==0){  sum++; x=i;  }
        }
        if(sum==1){
            sum=0;
            for(int i=1;i<=n;i++){
                if(low[i]==x) sum++;
            }
            printf("%d\n",sum);
        }
        else puts("0");
    }
    return 0;
}