HDU 5762

Teacher Bo

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1052    Accepted Submission(s): 582

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

 

Author

绍兴一中

 

Source

2016 Multi-University Training Contest 3

 

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题意：

给出若干组点的坐标，求这中间有没有两组曼哈顿距离相同的坐标。

思路：

题目中X，Y都小于M，就可以用每一种曼哈顿距离作为数组的下标，并把数组标记为1，暴力出数组值为1的那个下标值，就找出了相同的。

 代码：

 #include<iostream>
 #include<string>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<iomanip>
 #include<queue>
 using namespace std;
 struct point
 {
     int x,y;
 };
 int main()
 {
     int n,m,t;
     cin>>t;
     while(t--)
     {
         point p[];
         int dis[];
         cin>>n>>m;
         for(int i=;i<n;i++)
         {
             cin>>p[i].x>>p[i].y;
         }
         int flag=;
         memset(dis,,sizeof(dis));
         for(int i=;i<n;i++)
         {
             for(int j=i+;j<n;j++)
             {
                 int tem=abs(p[i].x-p[j].x)+abs(p[i].y-p[j].y);
                 if(dis[tem]==)
                 {
                     flag=;
                     break;
                 }
                 dis[tem]=;
             }
             if(flag==)
             {
                 cout<<"YES"<<endl;
                 break;
             }
         }
         if(flag==) cout<<"NO"<<endl;
     }
     return ;
 }