poj2002 哈希

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 17666		Accepted: 6735

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
先枚举两个点，通过数学公式得到另外2个点，使得这四个点能够成正方形。然后检查散点集中是否存在计算出来的那两个点，若存在，说明有一个正方形。
但这种做法会使同一个正方形按照不同的顺序被枚举了四次，因此最后的结果要除以4.

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)

可以用向量坐标来证明 对角线上俩坐标已知求另一条对角线坐标
标记点x y时，key = (x^2+y^2)%prime
解决的地址冲突的方法，我使用了 [b]链地址法

[/b]

#include<iostream>  //1500K	 1000MS
#include<cstdio>
#include<cstring>
#include<cmath>
#define F 19999

using namespace std;

struct zuo
{
int x,y;
} p[20001];
struct node
{
int x,y;
node *next;
}*head[20001];
int n;
int KK(zuo p1)
{
int key=(p1.x*p1.x+p1.y*p1.y)%F;
return key;
}
int Build(int k)  //建立
{
int key=KK(p[k]);
if(!head[key])
{
head[key]=new node;
head[key]->next=NULL;
node *q;
q=new node;
q->x=p[k].x;
q->y=p[k].y;
q->next=NULL;
head[key]->next=q;
}
else
{
node *q,*top;
top=head[key];
q=head[key]->next;
while(q)
{
q=q->next;
top=top->next;
}
q=new node;
q->next=NULL;
q->x=p[k].x;
q->y=p[k].y;
top->next=q;
}
return 0;
}
int Count(zuo p1,zuo p2)  //统计
{
int key1=KK(p1);
int flag=0;
int key2=KK(p2);
if(head[key1]&&head[key2]) //判断p1,p2是否在哈希表里
{
node *q=head[key1];
while(q)
{
if(q->x==p1.x&&q->y==p1.y)
{
flag=1;
break;
}
q=q->next;
}
if(flag==0)
return 0;
else
{
node *q=head[key2];
while(q)
{
if(q->x==p2.x&&q->y==p2.y)
{
return 1;
}
q=q->next;
}
}
}
return 0;
}
int main()
{
while(~scanf("%d",&n))
{
memset(head,0,sizeof(head));
if(!n)
break;
for(int i=0; i<n; i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
Build(i);
}
int num=0;
for(int i=0; i<n; i++)
{
for(int j=i+1; j<n; j++)
{
zuo p1,p2;
p1.x=p[i].x+(p[i].y-p[j].y);
p1.y=p[i].y-(p[i].x-p[j].x);
p2.x=p[j].x+(p[i].y-p[j].y);
p2.y=p[j].y-(p[i].x-p[j].x);
num+=Count(p1,p2);

p1.x=p[i].x-(p[i].y-p[j].y);
p1.y=p[i].y+(p[i].x-p[j].x);
p2.x=p[j].x-(p[i].y-p[j].y);
p2.y=p[j].y+(p[i].x-p[j].x);
num+=Count(p1,p2);
}
}
printf("%d\n",num/4);
}
}