hdu 4324 Triangle LOVE

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2588    Accepted Submission(s): 1053

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

   
   
   
   

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

 

Sample Output

   
   
   
   

    
    
    
    Case #1: Yes
Case #2: No
   
   
   
   

 

/*题解： 
    拓扑排序问题，本题仅仅判断是否成环即可。 
*/

#include<cstdio>
#include<cstring>
int map[2002][2002],d[2002];
char s[2002];
int main()
{
    int t,n,i,j,k=0;
    scanf("%d",&t);
    while(t--)
    {
        memset(map,0,sizeof(map));
        memset(d,0,sizeof(d));
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
            scanf("%s",s);
            for(j=0; j<n; j++)
            {
                if(s[j]=='1')
                {
                    map[i][j]=1;
                    ++d[j];//统计入度 
                }
            }
        }
        for(i=0; i<n; i++)
        {
            int flag=-1;
            for(j=0; j<n; j++)
            {
                if(d[j]==0)
                {
                    flag=j;
                    d[j]--;//不要遗漏这一步 
                    break;
                }
            }
            if(flag==-1)
                break;
            for(j=0; j<n; j++)
            {
                if(map[flag][j])//删去与入度0连接的边 
                {
                    d[j]--;
                    map[flag][j]=0;
                }
            }
        }
        if(i<n)
            printf("Case #%d: Yes\n",++k);
        else
            printf("Case #%d: No\n",++k);
    }
    return 0;
}