HDU 5762 Teacher Bo （暴力）

Teacher Bo

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5762

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

Source

2016 Multi-University Training Contest 3

##题意：

给出平面上的N个点，求是否一个四元组(A,B,C,D)满足A\B之间的曼哈顿距离等于C\D之间距离.
条件(A

##题解：

这个题让我懵比的两个小时...
一开始先把曼哈顿距离想成了abs(dx+dy)，然后可以只考虑每个点的坐标和，从而误把题目转化为求一个数组中是否有两对数的和相等.
由于规模是1e5,然后一直在想O(nlogn)的算法...

首先，曼哈顿距离的正确定义是abs(dx)+abs(dy).
其次，题目的数据规模确实不能跑O(n^2)的算法，但是题目限制了坐标的范围为[0,1e5], 这意味着任意两点的曼哈顿距离只可能有2e5种可能，暴力最多只会判断2e5个组合，所以不会超时.
所以直接暴力判断所有组合即可，当出现重复的距离时停止判断(先离线数据).

这个题懵比这么久真的是非常不应该，所以以后卡题的时候，一定要重新读题和尝试转换思考角度.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 251000
#define mod 100000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n,m;

struct point {

int x,y;

}p[maxn];

bool dis[maxn];

int get_ans(point a, point b) {

return abs(a.x-b.x) + abs(a.y-b.y);

}

int main(int argc, char const *argv[])

{

//IN;

int t; cin >> t;
while(t--)
{
    memset(dis, 0, sizeof(dis));
    cin >> n >> m;

    int flag = 0;

    for(int i=1; i<=n; i++) {
        int x,y; scanf("%d %d", &x,&y);
        p[i].x=x; p[i].y=y;
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<i; j++) {
            int tmp = get_ans(p[i],p[j]);
            if(dis[tmp]) {
                flag = 1;
                break;
            }
            dis[tmp] = 1;
        }
        if(flag) break;
    }

    if(flag) puts("YES");
    else puts("NO");
}

return 0;

}