Triangle LOVE 杭电HDU4324 【拓扑排序】

最新推荐文章于 2019-02-15 19:53:27 发布

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本文介绍了一种通过图论中的拓扑排序算法来检测特定规模人群中是否存在‘三角恋’关系的方法。该算法读取人群之间的关系矩阵，并确定是否存在一种情况，即三人之间形成相互爱慕的闭合环。

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

Case #1: Yes
Case #2: No

//TLE了N次，发现数组开小了

#include<stdio.h>
#include<string.h>
char map[2020][2020];
int indegree[2020];
int n;
bool Topo_Sort()
{
	for(int i=1;i<=n;++i)
	{
		int temp=-1;
		for(int j=0;j<n;++j)
		{
			if(indegree[j]==0)
			{
				temp=j;
				break;
			}
		}
		if(temp==-1) return 1;
		indegree[temp]=-1;
		for(int j=0;j<n;++j)
		{
			if(map[temp][j])
			{
				indegree[j]--;
			}
		}
	}
	return 0;
}
int main()
{
	int T;
	int i,j;
	int num=0;
	scanf("%d",&T);
	while(T--)
	{
		memset(indegree,0,sizeof(indegree));
		scanf("%d",&n);

		for(i=0;i<n;++i)
		{
			scanf("%s",&map[i]);
			for(j=0;j<n;++j)			
				if(map[i][j]=='1')				
					indegree[j]++;
		}
		
		printf("Case #%d: ",++num);
		if(Topo_Sort()) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}