Squares(POJ--2002

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.
题意：多组输入，每组输入n表示有n个点，求这n个点总共能构成几个正方形。
思路：根据数学知识可知，只要知道任意两点就可以根据公式求出其他两点，
公式：已知： (x1,y1) (x2,y2)

则： x3=x1+(y1-y2) y3= y1-(x1-x2) x4=x2+(y1-y2) y4= y2-(x1-x2)

或 x3=x1-(y1-y2) y3= y1+(x1-x2) x4=x2-(y1-y2) y4= y2+(x1-x2)。

如果只是简单的暴力去枚举所有的点肯定会超时，则这时就要想到用哈希表去记录已经出现过的坐标，然后每 枚举两个点，求出以该两点得到的其他两点，看这两点是否存在就可知该四个点能不能构成一个正方形，由于 可能会出现重复的正方形所以要最后得数要除以2。

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

<span style="font-size:18px;">#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 2000
#define H 1999
using namespace std;
//int x
,y
;
typedef struct node
{
int x,y,next;              //next是为了防止存储发生冲突，当发生冲突时就另外再开辟新的空间去存储新的一点的信息，next是来记录新空间的位置的
} Node;
Node node
,nade
;
int n,cnt,hashtable[H];
long sum;
bool cmp(Node a,Node b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
void init()                  //初始化，让没有出现过的点全部都等于-1
{
for(int i=0; i<H; i++)
hashtable[i]=-1;
cnt=0;
sum=0;
}
void insertt(int x,int y)
{
int h=(x*x+y*y)%H;            //为了减少冲突，用平方取余去存储
node[cnt].x=x;
node[cnt].y=y;
node[cnt].next=hashtable[h];         //相当于逆序建表，如果发生冲突了，就在原来出现的位置插入一个新空间
hashtable[h]=cnt;
cnt++;
}
int serch(int x,int y)
{
int h=(x*x+y*y)%H;
int next;
next=hashtable[h];
while(next!=-1)
{
if(node[next].x==x&&node[next].y==y)         //如果已经找到的话就返回真值
return 1;
next=node[next].next;                                      //如果当前不是要找的点就接着往下找
}
return 0;
}
int main()
{
//freopen("oo.text","r",stdin);
int n,x3,x4,y3,y4,i,j;
while(~scanf("%d",&n)&&n)
{
init();
for(i=0; i<n; i++)
{
scanf("%d %d",&nade[i].x,&nade[i].y);
insertt(nade[i].x,nade[i].y);
}
sort(nade,nade+n,cmp);             //要记得给所有的点排序，如果不排序那么只枚举一遍最后结果除以2是不对的，可能会漏情况，如果不排序，就要用两个公式枚举两遍，然后结果除以4
for(i=0; i<n; i++)
{
for( j=i+1; j<n; j++)
{
x3=nade[i].x+(nade[j].y-nade[i].y);
y3=nade[i].y+(nade[i].x-nade[j].x);
x4=nade[j].x+(nade[j].y-nade[i].y);
y4=nade[j].y+(nade[i].x-nade[j].x);
if(serch(x3,y3)&&serch(x4,y4))
sum++;
}
}
sum>>=1;
printf("%ld\n",sum);
}
return 0;
}</span>