2016 Multi-University Training Contest 3-1011.Teacher Bo，暴力！

Teacher Bo

                                                        Time
Limit: 4000/2000 MS (Java/Others)    

                                                   Memory Limit: 131072/131072 K (Java/Others)

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

这题实在找不到什么方法做了，只能暴力试试，可是居然卡在曼哈顿路径的概念上了；

我们以为是两点间的距离公示，然后用hypot（ ，）WA了两遍，之后两小时一直像睡着了了一样；

早知道百度一下概念，也不会耽误这么多时间，最后两分钟改了求距离那里直接交一发A了；

#include<bits/stdc++.h>
using namespace std;
const int N=100000+10;
struct node
{
    int x,y;
} a[N];
int cmp(node a,node b)
{
    if(a.x!=b.x) return a.x<b.x;
    return a.y<b.y;
}
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        set<double>m;
        for(int i=0; i<n; i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        sort(a,a+n,cmp);
        for(int i=1; i<n; i++)
        {
            if(a[i].x==a[i-1].x&&a[i].y==a[i-1].y)
            {
                for(int j=i+1; j<n; j++)
                    a[j]=a[j-1];
                n--;
                i--;
            }
        }
        int f=0;
        for(int i=0; i<n; i++)
        {
            if(f) break;
            for(int j=i+1; j<n; j++)
            {
                double dis=abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y);//主要还是概念没搞清，其他地方全正确；
                if(m.find(dis)!=m.end())
                {
                    f=1;
                    break;
                }
                m.insert(dis);
            }
        }
        if(f) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

同！志！仍！需！努！力！