hdu 4324 Triangle LOVE

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 894    Accepted Submission(s): 373

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

Zhaozhouyang

 

Source

2012 Multi-University Training Contest 3

 

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//就是用不恢复标记、判断某2点距离是否相差2且有线相连
#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <queue>

#include <vector>

using namespace std;

char b[2001][2001];

bool f[2001],flag;

int dis[2001],n;

void dfs(int df)

{

    if(flag) return;

    f[df]=1;

    int i;

    for(i=1;i<=n;i++)

      if(b[df][i]=='1')

      {

         if(f[i]&&dis[df]==dis[i]+2)//判断是否成3角恋关系

         {

           flag=1;

           break;

         }

          else if(!f[i])

          {

            dis[i]=dis[df]+1;

            dfs(i);

          }

      }

}

int main()

{

    int t,cnt=1;

    int i;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        getchar();

        for(i=1;i<=n;i++)

         {    f[i]=0;

             scanf("%s",b[i]+1);

         }

          flag=0;dis[1]=0;

          dfs(1);

          printf("Case #%d: ",cnt++);

          if(flag) printf("Yes\n");

          else printf("No\n");

    }

    return 0;

}