HDU 5762:曼哈顿距离

最新推荐文章于 2020-08-17 14:21:43 发布

Eira_H

最新推荐文章于 2020-08-17 14:21:43 发布

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分类专栏： ACM

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本文介绍了一道关于曼哈顿距离的算法题目，主要内容是如何判断在给定点集中是否存在两对不同的点，使得它们之间的曼哈顿距离相等。通过哈希表的应用，有效地解决了这一问题，并附上了C++实现的代码。

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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

Sample Output

    
     YES
NO

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;
}