poj 2186 Popular Cows ：求能被有多少点是能被所有点到达的点 tarjan O(E)

/**
problem: http://poj.org/problem?id=2186

当出度为0的点（可能是缩点后的点）只有一个时就存在被所有牛崇拜的牛
因为如果存在有两个及以上出度为0的点的话，他们不会互相崇拜所以不存在被所有牛崇拜的牛
牛的个数即该出度为0点（由环缩点而来）有多少牛
**/
#include<stdio.h>
#include<stack>
#include<algorithm>
using namespace std;

class Graphics{
    const static int MAXN = 10005;
    const static int MAXM = 50005;
private:
    struct Edge{
        int to, next;
    }edge[MAXM];
    struct Point{
        int dfn, low, color;
    }point[MAXN];
    int sign, dfnNum, colorNum, sumOfPoint, first[MAXN], count[MAXN];
    bool vis[MAXN];
    stack<int> stk;
    void tarjan(int u){
        point[u].dfn = ++dfnNum;
        point[u].low = dfnNum;
        vis[u] = true;
        stk.push(u);
        for(int i = first[u]; i != -1; i = edge[i].next){
            int to = edge[i].to;
            if(!point[to].dfn){
                tarjan(to);
                point[u].low = min(point[to].low, point[u].low);
            }else if(vis[to]){
                point[u].low = min(point[to].dfn, point[u].low);
            }
        }
        if(point[u].dfn == point[u].low){
            vis[u] = false;
            point[u].color = ++colorNum;
            count[colorNum] ++;
            while(stk.top() != u){
                vis[stk.top()] = false;
                point[stk.top()].color = colorNum;
                count[colorNum] ++;
                stk.pop();
            }
            stk.pop();
        }
    }
public:
    void clear(int n){
        sign = dfnNum = colorNum = 0;
        for(int i = 1; i <= n; i ++){
            first[i] = -1;
            vis[i] = 0;
            count[i] = 0;
        }
        sumOfPoint = n;
        while(!stk.empty()) stk.pop();
    }
    void addEdgeOneWay(int u, int v){
        edge[sign].to = v;
        edge[sign].next = first[u];
        first[u] = sign ++;
    }
    void addEdgeTwoWay(int u, int v){
        addEdgeOneWay(u, v);
        addEdgeOneWay(v, u);
    }
    void tarjanAllPoint(){
        for(int i = 1; i <= sumOfPoint; i ++){
            if(!point[i].dfn)
                tarjan(i);
        }
    }
    int getAns(){
        int ans = 0, ans2 = 0;
        int *outdegree = new int[sumOfPoint+1];
        for(int i = 1; i <= sumOfPoint; i ++){
            outdegree[i] = 0;
        }
        tarjanAllPoint();
        for(int i = 1; i <= sumOfPoint; i ++){
            for(int j = first[i]; j != -1; j = edge[j].next){
                int to = edge[j].to;
                if(point[to].color != point[i].color){
                    outdegree[point[i].color] ++;
                }
            }
        }
        for(int i = 1; i <= colorNum; i ++){
            if(!outdegree[i]){
                ans ++;
                ans2 = count[i];
            }
        }
        delete []outdegree;
        if(ans == 1){
            return ans2;
        }else{
            return 0;
        }
    }
}graph;

int main(){
    int n, m;
    scanf("%d%d", &n, &m);
    graph.clear(n);
    while(m --){
        int a, b;
        scanf("%d%d", &a, &b);
        graph.addEdgeOneWay(a, b);
    }
    printf("%d
", graph.getAns());
    return 0;
}