HDU4324 Triangle Love【拓扑排序】

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本文介绍了一道关于判断特定人群是否存在三角恋爱关系的编程竞赛题目。通过输入的人际关系矩阵，利用图论中的概念来解决该问题。文章提供了完整的算法思路及C++实现代码。

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Triangle LOVE

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 48 Accepted Submission(s) : 30

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[sub]i,j[/sub] = 1 means i-th people loves j-th people, otherwise A[sub]i,j[/sub] = 0. It is guaranteed that the given relationship is a tournament, that is, A[sub]i,i[/sub]= 0, A[sub]i,j[/sub] ≠ A[sub]j,i[/sub](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

Case #1: Yes
Case #2: No
【代码】
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define N 2010 
int n,indegree[N],flag;
char str[N][N];
int main()
{
	int t,k=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		int flag=0;
		memset(indegree,0,sizeof(indegree));
		int i,j;
		for(i=0;i<n;i++)
		{
			scanf("%s",str[i]);
			for(j=0;j<n;j++)
			    if(str[i][j]=='1')
			        indegree[j]++;
		}
		for(i=0;i<n;i++)
	    {
		    for(j=0;j<n;j++)
	            if(indegree[j]==0)
	                break;
	        if(j==n)//任何一个节点入度都不为0,说明存在环 
	        {
	    	    flag=1;
	    	    break;
		    }
		    else
	 	    {
			    indegree[j]--;
			    for(int p=0;p<n;p++)
			        if(str[j][p]=='1')
			            indegree[p]--;
		    }
	    }
		if(flag)
		    printf("Case #%d: Yes\n",++k);
		else
		    printf("Case #%d: No\n",++k);    
		    
	}
	return 0;
}