hdu4324

http://acm.hdu.edu.cn/showproblem.php?pid=4324

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
   
   
   
   

/**
hdu4324  dfs

题目大意：
         判断是否存在点数为3的环。
解题思路:
         dfs搜索即可，不做标记会TLE，dfs时做标记如果当前点u的下一个点v已经访问过了，
         那么就判断dist[u]==dist[[v]+2,成立返回true,否则更新dist[v]=dist[u]+1,继续深搜。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

char s[2005][2005];
int n,dist[2005];

struct note
{
    int v,next;
} edge[2005*2005];

int head[2005],ip,flag[2005];

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

void addedge(int u,int v)
{
    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;
}

bool dfs(int u)
{
    flag[u]=1;
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].v;
        if(flag[v]&&dist[u]==dist[v]+2)
        {
            puts("Yes");
            return 1;
        }
        else if(flag[v]==0)
        {
            dist[v]=dist[u]+1;
            if(dfs(v))
                return 1;
        }
    }
    return 0;
}

int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%s",s[i]);
        init();
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(s[i][j]=='1')
                {
                    addedge(i+1,j+1);
                }
            }
        }
        memset(flag,0,sizeof(flag));
        memset(dist,0,sizeof(dist));
        int flag1=0;
        printf("Case #%d: ",++tt);
        for(int i=1; i<=n; i++)
        {
            if(dfs(i)==1)
            {
                flag1=1;
                break;
            }
        }
        if(flag1==0)
            puts("No");
    }
    return 0;
}