POJ-2002 Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 18416		Accepted: 7081

Description
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with
the latter property, however, as a regular octagon also has this property.

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x
and y coordinates.

Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each
point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input

4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0

Sample Output

1
6
1

Source
Rocky Mountain 2004

分析： 暴力枚举两点（对角线或一边），然后用hash判断另外两点是否存在，向量旋转公式为：[x*cosA-y*sinA x*sinA+y*cosA] 。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
struct POINT
{
int x,y;
} point[1001],f[117007];
bool jud[117007];
int n,ans;
void insert(int x,int y)
{
int k=(abs(x)*10+abs(y))%107007;
while(jud[k] && (f[k].x != x || f[k].y != y)) k++;
if(!jud[k])
{
jud[k]=true;
f[k].x=x;
f[k].y=y;
}
}
bool Find(int x,int y)
{
int k=(abs(x)*10+abs(y))%107007;
while(jud[k] && (f[k].x != x || f[k].y != y)) k++;
if(!jud[k]) return false;
return true;
}
int main()
{
cin.sync_with_stdio(false);
while(cin >> n && n)
{
ans=0;
memset(jud,0,sizeof(jud));
for(int i=1;i <= n;i++)
{
cin>>point[i].x>>point[i].y;
insert(point[i].x,point[i].y);
}
for(int i=1;i <= n;i++)
for(int j=1;j <= n;j++)
if(point[j].x > point[i].x && point[j].y > point[i].y)
{
int a=point[i].x,b=point[i].y,x=point[j].x,y=point[j].y;
if((x-a == y-b) && Find(a,y) && Find(x,b)) ans++;
int dx=b-y,dy=x-a;
if(Find(x+dx,y+dy) && Find(a+dx,b+dy)) ans++;
}
cout<<ans<<endl;
}
}