hdu4324——Triangle LOVE

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

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 

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简单题，判断是否有环，拓扑排序搞定

#include
#include
#include
#include
#include

using namespace std;

const int N=2005;

struct node
{
    int next;
    int to;
}edge[N*N];
int head[N];
int tot,n;
char str[N];
int in_degree[N];

void addedge(int from,int to)
{
    edge[tot].to=to;
    edge[tot].next=head[from];
    head[from]=tot++;
}

void init()
{
    memset(head,-1,sizeof(head));
    memset(in_degree,0,sizeof(head));
    tot=0;
}

void topo_sort()
{
	int cnt=0;
    queuequ;
    while (!qu.empty())
        qu.pop();
    for (int i=1;i<=n;i++)
        if (!in_degree[i])
            qu.push(i);
    while (!qu.empty())
    {
        int u=qu.front();
        qu.pop();
        cnt++;
        for (int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            in_degree[v]--;
            if(!in_degree[v])
            	qu.push(v);
        }
        
    }
	if(cnt==n)
		printf("No\n");
	else
		printf("Yes\n");
}

int main()
{
	int t;
	int icase=1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		init();
		for(int i=1;i<=n;i++)
		{
			scanf("%s",str);
			for(int j=0;j