杭电5762之Teacher Bo_5762分解质因数-优快云博客

本文探讨了一道编程题目，该题要求判断在地图上是否存在四点，其中任意两点之间的曼哈顿距离相等。通过分析并提供了一个有效的算法解决方案，该方案利用了抽屉原理来优化计算效率。

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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：求是否存在四点（a,b,c,d），使得两点(a,b)与另外两点(c,d)间的曼哈顿距离相等

AC代码如下：
#include <iostream> 
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <map>
using namespace std;

#define mem(a) memset(a, 0, sizeof(a))
const int maxn = 1e5+100;
int b[maxn], q[2*maxn+100],a[maxn];


int main() {
    int t,i,n,m,j,f,res;
    scanf("%d",&t);
    while (t --) {
        mem(q);
        scanf("%d%d",&n,&m);    
        for (i = 1; i<=n; i++) 
        {
            scanf("%d%d",&a[i], &b[i]);    
        }
        if (n>1000)    printf("YES\n");//n等于1000时，有1000*999/2种点的组合（距离），如果有相同的距离组合，输出YES，若全部相同，
					//m最大值为1e5（距离为整数，故最多有2*1e5种距离），1000*999/2>1e5,
        else                            //由抽屉原理可知肯定有两个距离相等，故也应输出YES，1000仍可缩小，进一步减少运行时间，
					//此处节省运行时间，否则后面的代码枚举必定超时
        {
            res = 0;
            for (i = 1; i<=n; i++) 
            {
                for (j = i+1; j<=n; j++) 
                {
                    f = abs(a[j]-a[i])+abs(b[j]-b[i]);
                    q[f] ++ ;//保存距离不同的组合
                    if (q[f] > 1) //大于1,则存在相等距离
                    {
                        res = 1;
                        goto end;//退出双层循环，至 下面的 end: 位置
                    }
                }
            }
end:
            if (res == 0)    printf("NO\n");
            else    printf("YES\n");
        }
        
    }
    return 0;
}