Triangle LOVE
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Status
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). 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).
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 : Grant Yuan
time : 2014.7.29
algorithm: topological_sort
source : HDU 4324
notice : n<3时输出为no
**********************************************/
#include
#include
#include
#include
#include
#define MAX 2005 using namespace std; char mat[MAX][MAX]; int num[MAX]; int n,m,ans,sum; int lu[MAX]; int ct=1; int main() { int t; scanf("%d",&t); while(t--){ memset(mat,0,sizeof(mat)); memset(num,0,sizeof(num)); memset(lu,0,sizeof(lu)); scanf(" %d",&n); for(int i=0;i
=0;i--) { if(num[i]==0) break; } if(i==-1) break; num[i]=-1; lu[p--]=i; for(int j=0;j
-1) printf("Case #%d: Yes\n",ct++); else printf("Case #%d: No\n",ct++);} } return 0; }
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