poj 2186 Popular Cows

题目链接：Popular Cows

题目大意：有n个奶牛，m个关系，如果a认为b是受欢迎的，b认为c是受欢迎的，那么a认为c是受欢迎的，问有几个奶牛被其他所有的奶牛都认为是受欢迎的

题目思路：先建有向图，那么问题可以转化为有多少个点被其他所有点访问到，我们可以先缩点，然后看出度为0的连通块有多少个，只能有1个，否则不可能存在这样的奶牛，然后看这个出度为0的连通块里面有多少个点，输出就好

#include <map>#include <set>#include <cmath>#include <stack>#include <queue>#include <vector>#include <cstdio>#include <string>#include <utility>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 20010;vector<int>Edge[maxn];stack<int>S;int Dfn[maxn],Low[maxn],sccno[maxn],tclock,sccnt;int OutDeg[maxn],numscc[maxn],popu;void tarjan(int u){    Dfn[u] = Low[u] = ++tclock;    S.push(u);    for(int i = 0;i < Edge[u].size();i++){        int v = Edge[u][i];        if(!Dfn[v]){            tarjan(v);            Low[u] = min(Low[u],Low[v]);        }        else if(!sccno[v]){            Low[u] = min(Low[u],Dfn[v]);        }    }    if(Low[u] == Dfn[u]){        sccnt ++;        int v = -1;        while(v != u){            v = S.top();            S.pop();            sccno[v] = sccnt;            numscc[sccnt]++;        }    }}void findscc(int n){    tclock = sccnt = 0;    memset(Dfn,0,sizeof(Dfn));    memset(Low,0,sizeof(Low));    memset(sccno,0,sizeof(sccno));    memset(numscc,0,sizeof(numscc));    for(int i = 1;i <= n;i++)        if(!Dfn[i]) tarjan(i);}int solve(int n){    memset(OutDeg,0,sizeof(OutDeg));    popu = 0;    for(int u = 1;u <= n;u++){        for(int i = 0;i < Edge[u].size();i++){            int v = Edge[u][i];            if(sccno[u] != sccno[v]){                OutDeg[sccno[u]]++;            }        }    }    int c = 0;    for(int i = 1;i <= sccnt;i++){        if(OutDeg[i] == 0){            if(c == 0) {popu = numscc[i];c++;}            else popu = 0;        }    }    return popu;}int main(){    int T,n,m;    while(~scanf("%d%d",&n,&m)){        for(int i = 1;i <= n;i++) Edge[i].clear();        while(m--){            int u,v;            scanf("%d%d",&u,&v);            Edge[u].push_back(v);        }        findscc(n);        printf("%d\n",solve(n));    }    return 0;}