杭电4324Triangle LOVE三角恋

最新推荐文章于 2019-04-22 19:16:54 发布

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最新推荐文章于 2019-04-22 19:16:54 发布

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分类专栏：拓扑排序文章标签：拓扑排序杭电

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Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3508 Accepted Submission(s): 1364

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

Author

判断三角恋，第一行的意思就是0喜欢1,2,3,4；

第二行是1喜欢0；

第三行就是3喜欢1和4，依次类推：

附ac代码：

#include<stdio.h>
#include<string.h>
int topo[2020],i,n,m,j,k,l,t,map[2020][2020];
char c[2020][2020];
int main()
{
	int biaoji=1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(topo,0,sizeof(topo));
		for(i=0;i<n;i++)
		scanf("%s",&c[i]);
		for(i=0;i<n;i++)
		for(j=0;j<n;j++)
		{
			if(c[i][j]=='1')
			topo[j]++;
		}
		int flag=0;
		for(i=0;i<n;i++)
		{
			if(topo[i]==0)
			{
				topo[i]=1000;//经调试，如果i有三角恋关系类似的
				for(j=0;j<n;j++)//则topo[i]一定减不到0
				if(c[i][j]!='0')
				topo[j]--;
				flag++;//则flag+不到n
				i=0;
			}
		}
		printf("Case #%d: ",biaoji++);
		if(flag==n)
		printf("No\n");
		else
		printf("Yes\n");
	}
	return 0;
}