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

 

Sample Output

 

Author

BJTU

 

Source

 

Recommend

zhoujiaqi2010

 

题意分析(转载)：

此题可以一遍拓扑排序判环求解 即只需要找到一个环，

就必定存在三元环 证明如下： 假设存在一个n元环，

因为a->b有边，b->a必定没边，反之也成立

所以假设有环上三个相邻的点a-> b-> c,那么如果c->a间有边，

就已经形成了一个三元环，如果c->a没边，那么a->c肯定有边，

这样就形成了一个n-1元环。。。。

所以只需证明n大于3时一定有三元环即可，显然成立。

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

const int N=;

int n,indeg[N];     //存储的是节点的入度

char str[N][N];

int main(){

    //freopen("input.txt","r",stdin);

    int t,cases=;

    scanf("%d",&t);

    while(t--){

        scanf("%d",&n);

        int flag=;

        memset(indeg,,sizeof(indeg));  //将所有的节点入度初始化为0

        int i,j;

        for(i=;i<n;i++){

            scanf("%s",str[i]);

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

                if(str[i][j]=='')  //如果i喜欢j,则把j的入度加1

                    indeg[j]++;

        }

        for(i=;i<n;i++){

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

                if(indeg[j]==)     //找出入度为0的节点

                    break;

            if(j==n){       //任何一个节点的入度都不为0，说明存在环了，则必有三角恋

                flag=;

                break;

            }else{

                indeg[j]--; //除去当前结点

                for(int k=;k<n;k++)    //把从这个节点出发的引起的节点的入度都减去1

                    if(str[j][k]=='')

                        indeg[k]--;

            }

        }

        if(flag)

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

        else

            printf("Case #%d: No\n",++cases);

    }

    return ;

}