POJ 2002 Squares

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 18718		Accepted: 7209

Description
A square is a 4-sided polygon(多边形) whose sides have equal length and adjacent(邻近的) sides
form 90-degree angles. It is also a polygon such that rotating(旋转的) 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(八边形) 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(承担) that
the night sky is a 2-dimensional plane, and each star is specified(指定) by its x and y coordinates(坐标). 

Input
The input(投入) consists of a number of test cases. Each test case starts
with the integer(整数) n (1 <= n <= 1000) indicating(表明) 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(明显的) and
the magnitudes(大小) of the coordinates are less than 20000. The input is terminated(终止) 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

因为坐标有负数，所以在生成key值时要注意一下，思路很简单，先枚举两个点所有的组合，然后通过数学公式的换算来计算另外两个点，然后用哈希表查找一下那两个点是否存在。因为我们把所有点都枚举过一次，所以一个正方形被我们找过四次，把最后结果除以四就好了。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<map>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#include<new>

using namespace std;

const int mod = 100007;

struct node
{
int x;
int y;
}dian[1234];

node *hs[mod+6];

bool f(int xx,int yy,int xxx,int yyy)
{
long long int t;
int flag=0;
t=xx*xx+yy*yy;
t=t%mod+1;
while(hs[t])
{
if(hs[t]->x==xx&&hs[t]->y==yy)
{
flag++;
break;
}
t++;
}
t=xxx*xxx+yyy*yyy;
t=t%mod+1;
while(hs[t])
{
if(hs[t]->x==xxx&&hs[t]->y==yyy)
{
flag++;
break;
}
t++;
}
if(flag==2) return 1;
else return 0;
}

int main()
{
int n;
while(cin>>n&&n)
{
memset(hs,0,sizeof(hs));
for(int i=1;i<=n;i++)
{
scanf("%d%d",&dian[i].x,&dian[i].y);
long long int key=(dian[i].x*dian[i].x)+(dian[i].y*dian[i].y);
key=key%mod+1;
if(hs[key]) while(hs[key]) ++key;
// hs[key]=new node;
hs[key]=&dian[i];
}
int sum=0;
for(int i=2;i<=n;i++)
for(int j=1;j<i;j++)
{
int x1=dian[i].x,x2=dian[j].x,y1=dian[i].y,y2=dian[j].y,x3,x4,y3,y4;
x3=x1+(y1-y2);
x4=x2+(y1-y2);
y3=y1-(x1-x2);
y4=y2-(x1-x2);
if(f(x3,y3,x4,y4)) sum++;
x3=x1-(y1-y2);
x4=x2-(y1-y2);
y3=y1+(x1-x2);
y4=y2+(x1-x2);
if(f(x3,y3,x4,y4)) sum++;
}
cout<<sum/4<<endl;
}
return 0;
}