hdu4324 Triangle LOVE(拓扑排序)

题意：给一个图，问存不存在三个顶点构成的环

思路：直接建图然后看看能否拓扑排序，可以拓扑排序则必定满足图是有向的并且不存在环

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=2000+10;
int n;
vector<int> G[maxn];
int in[maxn];
bool topo()
{
queue<int> Q;
for(int i=0;i<n;i++)
if(!in[i]) Q.push(i);
int sum=0;
while(!Q.empty())
{
int u=Q.front(); Q.pop();
sum++;
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(--in[v]==0) Q.push(v);
}
}
return sum==n;
}
int main()
{
int T; scanf("%d",&T);
for(int kase=1;kase<=T;kase++)
{
scanf("%d",&n);
memset(in,0,sizeof(in));
for(int i=0;i<n;i++)
{
G[i].clear();
char str[maxn];
scanf("%s",str);
for(int j=0;j<n;j++)if(str[j]=='1')
{
G[i].push_back(j);
in[j]++;
}
}
printf("Case #%d: %s\n",kase,topo()?"No":"Yes");
}
return 0;
}

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