HDU 5762:曼哈顿距离

[align=left]Problem Description[/align]
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".
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[align=left]Input[/align]
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).
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[align=left]Output[/align]
T
lines, each line is "YES" or "NO".
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[align=left]Sample Input[/align]

2
3 10
1 1
2 2
3 3
4 10
8 8
2 3
3 3
4 4

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[align=left]Sample Output[/align]

YES
NO

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[align=left]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。[/align]

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[align=left]AC代码：[/align]

#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;
}

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