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).  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

 

Author

Zhaozhouyang

 

Source

2012 Multi-University Training Contest 3

 

Recommend

zhoujiaqi2010

//这题做的好奇怪、开始用vector进行邻接存储以某点出发可以到的点
//结果超时，而用for循环（i=1;i<=n;i++）居然187Ms
//网上说官方解题报告用“增量法",表示不知道、唉、见识太少
//这方法是最早想到的、结果一直被自己排除，东搞西搞最后还是回来用dfs
//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;
}