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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

先枚舉兩個點，通過數學公式得到另外2個點，使得這四個點可以成正方形。然後檢查散點集中是否存在計算出來的那兩個點，若存在，說明有一個正方形。

但這樣的做法會使同一個正方形依照不同的順序被枚舉了四次。因此最後的結果要除以4.

已知： (x1,y1) (x2,y2)

則： 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)

能夠用向量坐標來證明 對角線上倆坐標已知求還有一條對角線坐標

標記點x y時，key = (x^2+y^2)%prime

解決的地址沖突的方法，我使用了 鏈地址法

#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)  //統計
{
    int key1=KK(p1);
    int flag=0;
    int key2=KK(p2);
    if(head[key1]&&head[key2]) //推斷p1,p2是否在哈希表裏
    {
        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);
    }
}

poj2002 哈希