Triangle LOVE（拓扑排序）

Triangle LOVE

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)
Total Submission(s) : 40 Accepted Submission(s) : 27
[align=left]Problem Description[/align]
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”.

[align=left]Input[/align]
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[sub]i,j[/sub] = 1 means i-th people loves j-th people, otherwise A[sub]i,j[/sub] = 0. It is guaranteed that the given relationship is a tournament, that is, A[sub]i,i[/sub]= 0, A[sub]i,j[/sub] ≠ A[sub]j,i[/sub](1<=i, j<=n,i≠j).

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Case #1: Yes
Case #2: No

[align=left]Author[/align]
BJTU

[align=left]Source[/align]
2012 Multi-University Training Contest 3

#include<stdio.h>
#include<string.h>
char map[2010][2010];
int dre[2010];
int main()
{
int t,cot=0;
scanf("%d",&t);
while(t--)
{
int n,i,j,p;
//memset(map,0,sizeof(map));
scanf("%d",&n);
memset(dre,0,sizeof(dre));
getchar();
for(i=0;i<n;i++)
{
scanf("%s",&map[i]);
getchar();
for(j=0;j<n;j++)
if(map[i][j]=='1')
dre[j]++;
}
for(i=0;i<n;i++)
{
p=-1;
for(j=0;j<n;j++)
{
if(dre[j]==0)
{
dre[j]--;
p=j;
break;
}
}
if(p==-1)
break;
for(j=0;j<n;j++)
{
if(map[p][j]=='1')
{
dre[j]--;
map[p][j]='0';
}
}
}
printf("Case #%d: ",++cot);
if(i<n)
{
printf("Yes\n");
}
else
printf("No\n");
}
}