POJ 2186 强连通分量 Targin算法

        第一道强连通分量的题，，，泪牛满面啊，，，话说，看这个算法有4、5天了吧，今天终于写出来一道题，，纠结。几天写了一道题，，，这可怎么办？？？这道题就是先求出强连通分量的个数，如果强连通分量个数为1，则图是强连通图，输出n即可。否则的话，统计其中独立强连通分量的个数，所谓独立强连通分量，就是外面的点到不了该强连通分量。若独立强连通分量的个数为1，则输出出度为0的强连通分量所包含的点得个数，否则输出0.

题目：

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15106		Accepted: 5990

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

Hint

Cow 3 is the only cow of high popularity. 

ac代码：

#include <iostream>
#include <string.h>
#include <vector>
#include <cstdio>
using namespace std;
const int N=10010;
int dfn[N],in[N],out[N],belong[N];
int low[N],num[N];
vector<int> ss[N];
int stack[N],instack[N];
int top,index,cnt;
void init(){
  memset(dfn,0,sizeof(dfn));
  memset(in,0,sizeof(in));
  memset(out,0,sizeof(out));
  memset(belong,0,sizeof(belong));
  memset(low,0,sizeof(low));
  memset(num,0,sizeof(num));
  memset(ss,0,sizeof(ss));
  memset(stack,0,sizeof(stack));
  memset(instack,0,sizeof(instack));
}
int min(int x,int y){
  if(x<y)
	  return x;
  return y;
}
void targin(int x){
	int j;
  low[x]=dfn[x]=++index;
  stack[++top]=x;
  instack[x]=1;
  for(int i=0;i<ss[x].size();++i){    
    j=ss[x][i];
	if(dfn[j]==0)
	{targin(j);low[x]=min(low[x],low[j]);}
	else if(instack[j]){
	  low[x]=min(low[x],dfn[j]);
	}
  }
  if(dfn[x]==low[x]){
    cnt++;
	do{
	  j=stack[top--];
	  belong[j]=cnt;
	  num[cnt]++;
	  instack[j]=0;
	}while(x!=j);
  }
}
int main(){
	//freopen("2.txt","r",stdin);
  int n,m;
  while(~scanf("%d%d",&n,&m)){
    init();
	top=index=cnt=0;
	int x,y;
	while(m--){
	  scanf("%d%d",&x,&y);
	  ss[x].push_back(y);
	}
	for(int i=1;i<=n;++i){
	  if(dfn[i]==0)
		  targin(i);
	}
	/*for(int i=1;i<=n;++i)
		printf("%d ",low[i]);
	printf("\n");*/
	for(int i=1;i<=n;++i){
		for(int j=0;j<ss[i].size();++j){
		  x=belong[i];
		  y=belong[ss[i][j]];
		  if(x!=y){
		    out[x]=1;
			in[y]=1;
		  }
		}
	}
	//printf("cnt==%d\n",cnt);
	if(cnt==1)
	{
	  printf("%d\n",n);
	  continue;
	}
	int sum=0,count=0,pos;
	for(int i=1;i<=cnt;++i){
	  if(out[i]==0)
	  {count++; pos=i;}
	}
	//printf("count==%d\n",count);
	//printf("pos")
	if(count==1){
		for(int i=1;i<=n;++i)
			if(belong[i]==pos)
				sum++;
		printf("%d\n",sum);
	}
	else
		printf("0\n");
  }
  return 0;
}