HDU 5762:曼哈顿距离

Problem Description

Teacher BoBo is a geography teacher in the school.One day in his class,he markedN 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

23 101 12 23 34 108 82 33 34 4

 

Sample Output

YESNO

分析：这个题主要是要知道曼哈顿距离的意思，曼哈顿距离（Manhattan Distance）是由十九世纪的赫尔曼·闵可夫斯基所创词汇 ，是种使用在几何度量空间的几何学用语，用以标明两个点上在标准坐标系上的绝对轴距总和。然后hash一下就解决了。

AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<string.h>#include<algorithm>using namespace std;struct Point{    int x;    int y;};Point point[100010];bool ans[10000000];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int N,M;        scanf("%d%d",&N,&M);        for(int i=1;i<=N;i++)            scanf("%d%d",&point[i].x,&point[i].y);        memset(ans,0,sizeof(ans));        int flag=0;        for(int i=1;i<=N-1;i++)        {            for(int j=i+1;j<=N;j++)            {                int dis=abs(point[i].x-point[j].x)+abs(point[i].y-point[j].y);                if(ans[dis])                {                    flag=1;                    cout<<"YES"<<endl;                    break;                }                ans[dis]=true;            }            if(flag)                break;        }        if(flag==0)            cout<<"NO"<<endl;    }    return 0;}