Problem Description

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, A_i,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

题意：  有n个人 然后以下有n行（i） 每行有n个数字 假设第j个数字为1,表示i对j有好感。推断这些关系中是否有三角恋

代码：
#include <stdio.h>

#include <string.h>

int t,n;

//存储的是节点的入度

int in_degree[2010];

//存储的是i,j两个节点的关系，1:i love j,0:j love i

char adj_mat[2010][2010];

int main()

{

    bool flag;//true表示为有三角恋。false表示为没有三角恋

    scanf("%d",&t);

    for(int i = 1; i <= t;i++)

    {

        scanf("%d",&n);

        flag = false;

        //将全部的节点入度初始化为0

        memset(in_degree,0,sizeof(in_degree));

        for(int j = 0; j < n; j++)

        {

            scanf("%s",adj_mat[j]);

            for(int k=0;k<n;k++)

            if(adj_mat[j][k]=='1')//假设j喜欢k,则把k的入度加1

            in_degree[k]++;

        }

        for(int j=0;j<n;j++)

        {

            int k;

            for(k=0;k<n;k++)

            if(in_degree[k]==0)break;//找出入度为0的节点

            if(k==n)//不论什么一个节点的入度都不为0。说明存在环了，则必有三角恋

            {

                flag = true;

                break;

            }else{

                //将这个点的入度设为-1，避免再次循环时有查到了这个节点，

                //此时说明这个点已经从集合中除掉了

                in_degree[k]--;

                for(int p=0;p<n;p++)

                {

                    //把从这个节点出发的引起的节点的入度都减去1

                    if(adj_mat[k][p]=='1'&&in_degree[p]!=0)

                    in_degree[p]--;

                }

            }

        }

        if(flag)

        printf("Case #%d: Yes\n",i);

        else printf("Case #%d: No\n",i);

    }

    return 0;

}

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