HDU 4324 Triangle LOVE(拓扑排序)

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 4867    Accepted Submission(s): 1917


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). 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

BJTU

 

题意：给你喜欢关系，判断有没有三角恋。

思路：拓扑排序判断有没有环。

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int n;
char map[2001][2001];
int indegree[2001];
int cnt;
void toposort()
{
for(int i=1;i<=n;i++)
{
int flag=0;
for(int j=1;j<=n;j++)
{
if(indegree[j]==0)
{
cnt++;
flag=1;
indegree[j]--;
for(int k=1;k<=n;k++)
{
if(map[j][k]=='1')
indegree[k]--;
}
break;
}
}
if(!flag) break;
}
}
int main()
{
int t;
scanf("%d",&t);
for(int g=1;g<=t;g++)
{
scanf("%d",&n);
memset(indegree,0,sizeof(indegree));
for(int i=1;i<=n;i++)
{
scanf("%s",map[i]+1);
for(int j=1;j<=n;j++)
if(map[i][j]=='1')
indegree[j]++;
}
cnt=0;
toposort();
cout<<"Case #"<<g<<": ";
if(cnt==n) cout<<"No"<<endl;
else cout<<"Yes"<<endl;
}

return 0;
}