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 = ;
     const static int MAXM = ;
 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 != -; 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 = ;
         for(int i = ; i <= n; i ++){
             first[i] = -;
             vis[i] = ;
             count[i] = ;
         }
         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 = ; i <= sumOfPoint; i ++){
             if(!point[i].dfn)
                 tarjan(i);
         }
     }
     int getAns(){
         int ans = , ans2 = ;
         int *outdegree = new int[sumOfPoint+];
         for(int i = ; i <= sumOfPoint; i ++){
             outdegree[i] = ;
         }
         tarjanAllPoint();
         for(int i = ; i <= sumOfPoint; i ++){
             for(int j = first[i]; j != -; j = edge[j].next){
                 int to = edge[j].to;
                 if(point[to].color != point[i].color){
                     outdegree[point[i].color] ++;
                 }
             }
         }
         for(int i = ; i <= colorNum; i ++){
             if(!outdegree[i]){
                 ans ++;
                 ans2 = count[i];
             }
         }
         delete []outdegree;
         if(ans == ){
             return ans2;
         }else{
             return ;
         }
     }
 }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\n", graph.getAns());
     return ;
 }