Popular Cows

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
【题意】：给你一个有向图，问你有多少个点可以被其它所有点访问到。

我们可以用tarjia算法去掉所有的环，然后重构图后就变成了，找出出度为零的点就行了

https://www.byvoid.com/zhs/blog/scc-tarjan

https://blog.csdn.net/qq_34374664/article/details/77488976

/********************************
tarjan + 图的重构  
*********************************/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>

using namespace std;

const int maxn = 1e4+50;

stack<int>sta;
vector<int>G[maxn];
int dfn[maxn],low[maxn],tag[maxn];
int index,scc,id[maxn];
bool vis[maxn];
int n,m;

void init() {
    index = 0,scc = 0;
    memset(vis,false,sizeof(vis));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(tag,0,sizeof(tag));
    memset(id,0,sizeof(id));
    while(!sta.empty()) sta.pop();
    for (int i = 1;i<=n;i++) G[i].clear();
}

void tarjan(int x) {  //强联通tarjan 
    dfn[x] = low[x] = ++index;
    vis[x] = true; sta.push(x);
    int len = G[x].size();
    for (int i = 0 ; i < len;i++ ) {
        int to = G[x][i];
        if ( !vis[to] ) {
            tarjan(to);  low[x] = min(low[x],low[to]);
        } else if ( vis[to] ) {
            low[x] = min(low[x],dfn[to]);
        }
    }
    if(low[x] == dfn[x]) {  // ac代码,跟re代码区别不大关键是取栈顶的位置 
        scc++;
        int v;
        do{
            v = sta.top();
            tag[v] = scc;
            id[scc]++;
            sta.pop();
        }while(x != v);
    }
//  if(low[x] == dfn[x]) {   //re代码,栈越界 
//      scc++;
//      int v = sta.top();
//      do{
//          tag[v] = scc;
//          id[scc]++;
//          sta.pop();
//          v = sta.top();
//      }while(x!=v);
//  }
}

int main(){
    while(~scanf("%d %d",&n,&m)){
        int u,v;
        init();
        for (int i = 1 ; i <= m ; i ++) {
            scanf("%d %d",&u,&v);
            G[u].push_back(v);
        }
        for (int i = 1;i <= n; i ++ ) {
            if(!dfn[i]) tarjan(i);
        } 
        memset(vis,false,sizeof(vis));
        for (int i = 1;i <= n;i++ ) {  //利用tarjan算法缩点重构图 
            int len = G[i].size();
            for (int j = 0;j < len;j++) {
                    if(tag[i] != tag[G[i][j]] ) {
                    vis[tag[i]] = true;
                }
            }
        }
        int pos,sum = 0;
        for (int i = 1;i<=scc;i++) {
            if(!vis[i]) sum++,pos = i;
        }
        if(sum == 1) {
            printf("%d
",id[pos]);
        } else printf("0
");
    }
    return 0;
}
/*
3 3
1 2
2 1
2 3
*/