hdu 5762 Teacher Bo 曼哈顿路径

Teacher Bo

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1014    Accepted Submission(s): 561

Problem Description

Teacher BoBo is a geography teacher in the school.One day in his class,he marked N points in the map,the i-th point is at (Xi,Yi).He wonders,whether there is a tetrad (A,B,C,D)(A<B,C<D,A≠CorB≠D) such that the manhattan distance between A and B is equal to the manhattan distance between C and D.

If there exists such tetrad,print "YES",else print "NO".

 

 

Input

First line, an integer T. There are T test cases.(T≤50)

In each test case,the first line contains two intergers, N, M, means the number of points and the range of the coordinates.(N,M≤105).

Next N lines, the i-th line shows the coordinate of the i-th point.(Xi,Yi)(0≤Xi,Yi≤M).

 

 

Output

T lines, each line is "YES" or "NO".

 

 

Sample Input

2 3 10 1 1 2 2 3 3 4 10 8 8 2 3 3 3 4 4

 

 

Sample Output

YES NO

#include<iostream>
#include<stdio.h>
#include<set>
#include<math.h>
using namespace std;
const int maxx = 100005;
set<int> hav;
int px[maxx];
int py[maxx];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        hav.clear();
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++ )
        {
            scanf("%d%d",px+i,py+i);
        }
        int flag=0;
        for(int i=0; i<n-1; i++)
        {
            for(int j=i+1; j<n; j++)
            {
                int dis=abs(px[j]-px[i])+abs(py[j]-py[i]);
                if(hav.count(dis))
                {
                    flag=1;
                    break;
                }
                else
                {
                    hav.insert(dis);
                }
            }
            if(flag)
            {
                break;
            }
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

考虑一种暴力,每次枚举两两点对之间的曼哈顿距离,并开一个桶记录每种距离是否出现过,如果某次枚举出现了以前出现的距离就输 YESYES ,否则就输 NONO .

注意到曼哈顿距离只有 O(M)O(M) 种,根据鸽笼原理,上面的算法在 O(M)O(M) 步之内一定会停止.所以是可以过得.

一组数据的时间复杂度 O(\min{N^2,M})O(min{N^2​​ ,M}) .