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
题目大意：给你一个图，图中随意两点之间要么有正向边，要么有反向边。
推断是否含有a->b->c->a的三角形环。
思路：事实上仅仅要有环，就能构成三角形环。
由于随意两点之间要么有正向边，
要么有反向边。假设如今有一个四元素环 a->b->c->d->a，若a不指向c，则
c必然指向a，所以必然存在三角形环。直接拓扑排序，假设不能排序。则有
三角环，输出“Yes”，能拓扑排序。则不含有三角环，输出"No"。
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 2010; int N,M,t;
int topo[MAXN],G[MAXN][MAXN],vis[MAXN];
char Map[MAXN][MAXN]; bool dfs(int u)
{
vis[u] = -1;
for(int v = 0; v < N; v++)
{
if(G[u][v])
{
if(vis[v] < 0)
return false;
else if(!vis[v] && !dfs(v))
return false;
}
}
vis[u] = 1;
topo[--t] = u;
return true;
} bool toposort()
{
t = N;
memset(vis,0,sizeof(vis));
for(int u = 0; u < N; u++)
{
if(!vis[u])
if(!dfs(u))
return false;
}
return true;
} int main()
{
int T,kase = 0;
cin >> T;
while(T--)
{
memset(G,0,sizeof(G));
memset(topo,0,sizeof(topo));
getchar();
cin >> N;
for(int i = 0; i < N; i++)
cin >> Map[i];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
if(Map[i][j] == '1')
G[i][j] = 1; }
cout << "Case #" << ++kase << ": ";
if(toposort())
cout << "No" << endl;
else
cout << "Yes" << endl;
} return 0;
}