HDU 4324 Triangle LOVE

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3542    Accepted Submission(s): 1381

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

25001001000001001111011100050111100000010000110001110

 

Sample Output

Case #1: YesCase #2: No

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 

Recommend

zhoujiaqi2010   |   We have carefully selected several similar problems for you:  1068 4320 4325 3309 3712

//这道题就是用拓扑的方法来判断是否成环，我的做法就是看是否能够排成有序的序列，参考代码如下：

 

 

#include<stdio.h>#include<string.h>int indegree[10010];char mp[10010][10010];int main(){  int  k=1;   int  flag;   int  t,n;   scanf("%d",&t);   while(t--)    {    int i,j;         //int m;     scanf("%d",&n);     memset(indegree,0,sizeof(indegree)); flag=0; for(i=0;i<n;i++) {    scanf("%s",mp[i]);    for(j=0;j<n;j++)    if(mp[i][j]=='1')indegree[j]++;   }  //int  m; for(j=0;j<n;j++)  {   for(i=0;i<n;i++)       if(indegree[i]==0)   break;              if(i==n)   {     flag=1;       break;   }  else      {         indegree[i]=-1;   for(int k=0;k<n;k++)     {     if(mp[i][k]=='1'&&indegree[k]!=0)       indegree[k]--;     }      }  }     if(flag)  printf("Case #%d: Yes\n",k++);  else  printf("Case #%d: No\n",k++);    }return 0;}