Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3603 Accepted Submission(s): 1416


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

輸入一個矩陣，如果a[i][j]==1，意思是i喜歡j，問存不存在一個三角戀關系， 既拓撲排序判環，，要深入理解拓撲排序的原理 2015,8,14

 

#include
#include
#define M 2100
char mp[M][M];
int du[M];
int n,cas;
void topo(){
	int ok=0,i,k;
	for(int i=1;i<=n;i++){
		for(k=1;k<=n;k++)
			if(du[k]==0) //z找到入度為0的點 
				break;
		if(k==n+1){//如果不存在入度為0的點，那麼就是一定存在環 ，就有三角戀 
			ok=1;
			break;
		}else{
			du[k]--;//刪除這個點 
			for(int j=1;j<=n;j++){
				if(mp[k][j]=='1'&&du[j]!=0)
					du[j]--;
			}
		}
	}
	if(ok) printf("Case #%d: Yes\n",cas++);
	else printf("Case #%d: No\n",cas++);
}
int main(){
	int i,j,t;
	cas=1;
	scanf("%d",&t);
	while(t--){
		memset(du,0,sizeof(du));
		scanf("%d",&n);
		for(i=1;i<=n;i++){//按照注釋的輸入會超時 
			//getchar(); 
			scanf("%s",mp[i]+1);;
			for(j=1;j<=n;j++){
			//	scanf("%c",&mp[i][j]);
				if(mp[i][j]=='1')
					du[j]++;
			}
		}
		topo();
	}
	return 0;
}