2015
05-23

# Triangle LOVE

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”.

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). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,i(1<=i, j<=n,i≠j).

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). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,i(1<=i, j<=n,i≠j).

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

Case #1: Yes
Case #2: No

/**
[有向环] hdu 4324 Triangle LOVE

*/
#include <stdio.h>
#include <string.h>
int in[2001];
int main(){
int t,cas = 0,i,j,n;
scanf("%d",&t);
while(t--){
memset(in,0,sizeof(in));
scanf("%d\n",&n);
for(i = 0;i < n; ++i){
for(j = 0; j < n; ++j)
if(getchar() == '1')
in[j]++;
getchar();
}

for(i = 0; i < n; ++i)
in[n] += (in[i] != 0);
printf(in[n] == n ? "Case #%d: Yes\n" : "Case #%d: No\n",++cas);
}
return 0;
}