Triangle LOVE（拓扑排序）

Triangle LOVE

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

Total Submission(s) : 40   Accepted Submission(s) : 27

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

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

 

Sample Output

Case #1: Yes
Case #2: No

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

#include<stdio.h>
#include<string.h>
char map[2010][2010];
int dre[2010];
int main()
{
	int t,cot=0;
	scanf("%d",&t);
	while(t--)
	{
		int n,i,j,p;
		//memset(map,0,sizeof(map));
		scanf("%d",&n);
		memset(dre,0,sizeof(dre));
		getchar();
		for(i=0;i<n;i++)
		{
			scanf("%s",&map[i]);
			getchar();
			for(j=0;j<n;j++)
			if(map[i][j]=='1')
			dre[j]++;
		}
		for(i=0;i<n;i++)
		{
			p=-1;
			for(j=0;j<n;j++)
			{
				if(dre[j]==0)
				{
					dre[j]--;
					p=j;
					break;
				}
			}
			if(p==-1)
				break;
			for(j=0;j<n;j++)
			{
				if(map[p][j]=='1')
				{
					dre[j]--;
					map[p][j]='0';
				}
			}
		}
		printf("Case #%d: ",++cot);
		if(i<n)
		{
			printf("Yes\n");
		}
		else
			printf("No\n");
	}
}