HDU 4324 Triangle LOVE 拓扑排序

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;
}

版权声明：本文博客原创文章，博客，未经同意，不得转载。