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題意：n頭牛，每頭牛的心目中都有本身最歡迎的牛，歡迎可傳遞，若是a認爲b最受歡迎，b認爲c最受歡迎，那麼a認爲c也受歡迎。統計能出本身外能讓其它全部的牛認爲受歡迎的牛的數量題解：統計全部點到一個點均可達的數量。有向無環圖中全部的點至某個點可達，那麼這個點的出度必爲0也是惟一的出度爲0的點.用tarjan算法縮點，統計縮點後的圖中出度爲0的數量，若是數量>1，那麼不存在不然輸出該點的強連通份量點的個數。#include<stdio.h>#include<iostream>#include<stack>#include<vector>#include<string.h>using namespace std;int visit[10001];int dfn[10001],low[10001],head[10001],n,m,ret,ans,ti;int flag[10001];stack<int> st;vector<int> v[10001];int du[10001][2];struct lmx{ int u; int v; int next;};lmx a[50001];void add(int u,int v){ a[ret].u=u; a[ret].v=v; a[ret].next=head[u]; head[u]=ret++;}void tarjan(int s){ dfn[s]=low[s]=++ti; st.push(s); visit[s]=1; int k; for(int i=head[s];i!=-1;i=a[i].next) {  int k=a[i].v;  if(!dfn[k])   {   tarjan(k);   if(low[s]>low[k]) low[s]=low[k];  }  else if(visit[k]&&low[s]>dfn[k]) low[s]=dfn[k]; } if(dfn[s]==low[s]) {  ans++;  do{   k=st.top();   st.pop();   visit[k]=0;   flag[k]=ans;   v[ans].push_back(k);  }while(s!=k); }}int cnt(){ int i,cnt=0,pos; if(ans==1) return n; memset(du,0,sizeof(du)); for(i=0;i<m;i++) {    int u=flag[a[i].u],v=flag[a[i].v];    if(u==v) continue;       du[u][0]++;    du[v][1]++; } for(i=1;i<=ans;i++) {  if(du[i][0]==0) {cnt++;pos=i;} } if(cnt>1) return 0; return v[pos].size();}int main(){ int i;    while(scanf("%d %d",&n,&m)!=EOF) {  memset(visit,0,sizeof(visit));  memset(dfn,0,sizeof(dfn));  memset(head,-1,sizeof(head));  ans=ti=ret=0;        for(i=0;i<m;i++)  {   int a,b;   scanf("%d %d",&a,&b);   add(a,b);  }  for(i=1;i<=n;i++) v[i].clear();  for(i=1;i<=n;i++)  {   if(!dfn[i]) tarjan(i);  }  printf("%d\n",cnt()); } return 0;}