Triangle LOVE

Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world! 
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A. 
  Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”. 

Input

The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000). 
In the next N lines contain the adjacency matrix A of the relationship (without spaces). A i,j = 1 means i-th people loves j-th people, otherwise A i,j = 0. 
It is guaranteed that the given relationship is a tournament, that is, A i,i= 0, Ai,j ≠ A j,i(1<=i, j<=n,i≠j).

Output

For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”. 
Take the sample output for more details. 

Sample Input

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

Sample Output

Case #1: Yes
Case #2: No

测试样例分析：

#include <cstdio>
#include <vector>
#include <queue>
#include <string.h>
#define MAX 2001 

using namespace std;

int n,in[MAX];
vector <vector<int> > v;

bool toposort(){
    int cnt=0;
    queue <int> q;
    //初始化，把所有入度为0的点入队 
    for(int i=0;i<n;i++){
        if(in[i]==0) q.push(i);
    }
    
    while(!q.empty()){//当还有点未处理 
        int u=q.front(); q.pop(); //取出点并弹出队列 
        
        in[u]--;//in[u] = -1 标记该点已经弄过了 
        cnt++;//计数作用 
        
        for(int i=0;i<v[u].size();i++){
        	//将u所指向的所有点入度减一 
            int d=v[u][i];//d为所u指向的点 
            in[d]--;
            if(in[d]==0){//如果这个过程中，入度变为0，则入队 
                q.push(d);
            }
        }
    }
    return cnt!=n;
}
 
int main(){
    int t;//t组测试数据 
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
    	//初始化 
    	v.clear();
    	scanf("%d",&n);
    	v.resize(n);//resize()用于设置大小(size) 
    	memset(in,0,sizeof(in)); 
		
		//输入数据并用容量vector保存 
    	for(int j=0;j<n;j++){
    	    char st[MAX];
    	    scanf("%s",st);//输入一行数据 
    	    for(int k=0;k<n;k++){
        	    if(st[k]=='1'){
        	        //如果为1，就将k压入到j行中 
					v[j].push_back(k); //j-->k
        	        in[k]++;// k的入度加一 
            	}
        	}
   		}

        printf("Case #%d: ",i);
        
        bool hasCircle = toposort();
        
        if(hasCircle)
        	printf("Yes\n");
        else 
			printf("No\n");
			
    }
    return 0;
}