﻿HDU 4324 Triangle LOVE(拓扑排序)-Ocrosoft

# Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4423    Accepted Submission(s): 1755

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

Source

```#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <list>
#include <stack>
#include <string>
#include <vector>
#include <cstring>
#include <cctype>
#include <queue>
#include <set>
#include <climits>
#define ms(a) memset(a,0,sizeof(a))
using namespace std;
int n,t;
char mp[2001][2001];//邻接表保存边信息
int indegree[2001];//保存入度
queue<int> q;
bool toposort()
{
int cnt=0;
while(!q.empty())
{
t=q.front(),q.pop(),cnt++;
for(int i=0; i<n; i++)
if(mp[t][i]&&!(--indegree[i]))
q.push(i);
}
if(cnt==n)return 1;
return 0;
}
int main()
{
int N;
cin>>N;
for(int I=1; I<=N; I++)
{
cin>>n;
ms(indegree);
for(int i=0; i<n; i++)
{
scanf("%s",mp[i]);
for(int j=0;j<n;j++)
if(mp[i][j]=='1')indegree[j]++;
}
while(!q.empty())q.pop();
for(int i=0; i<n; i++)
if(!indegree[i])q.push(i);
bool ans=toposort();
printf("Case #%d: %s\n",I,ans?"No":"Yes");
}
return 0;
}```