Squares(哈希)

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 14328		Accepted: 5393

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
题意：给出n个点的坐标，计算这些点可以构成多少个正方形，不同顺序的相同四个点被视为同一个正方形。

思路：这里n最大是1000，显然一个点一个点的枚举不行。参考了别人的结题报告，据说有这样的定理：
　　　已知（x1,y1） 和 （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);
     因此可以先枚举两个点，根据这两个点（点1 ，点2）的坐标可以得到另外两个点（点3， 点4），若在哈希表中能找得到这两个（点3， 点4），说明能构成一个正方形，
     注意每个点被枚举了四次，最后要除以4；
     再者就是找哈希函数，这里用的平方取余法，每输入一个点，就将这个点插入哈希表中；
     这个题也受了 poj 3274的启发；

#include<stdio.h>
#include<string.h>

const int prime = 99991;
struct node
{
    int x,y;
}pos[1010];

struct HashTable
{
    int x;
    int y;
    struct HashTable* next;
}*Hash[prime];//Hash[]是指针数组，存放地址；
int n;

//插入哈希表
void hash_insert(int x, int y)
{
    int key = (x*x + y*y)%prime;//平方求余法；
    if(!Hash[key])
    {
        Hash[key] = new struct HashTable;
        Hash[key]->x = x;
        Hash[key]->y = y;
        Hash[key]->next = NULL;
    }
    else
    {
        struct HashTable *tmp = Hash[key];
        while(tmp->next)
            tmp = tmp->next;//开放寻址，直到next为空
        //插入新结点
        tmp->next = new struct HashTable;
        tmp->next->x = x;
        tmp->next->y = y;
        tmp->next->next = NULL;
    }
}
bool find(int x, int y)
{
    int key = (x*x+y*y)%prime;
    if(!Hash[key])
        return false;//key 对应的地址不存在，
    else
    {
        struct HashTable *tmp = Hash[key];
        while(tmp)
        {
            if(tmp->x == x && tmp->y == y)
                return true;
            tmp = tmp->next;
        }
        return false;
    }
}
int main()
{
    while(scanf("%d",&n)!= EOF)
    {
        int i,j;
        if(n == 0) break;
        memset(Hash,0,sizeof(Hash));
        for(i = 0; i < n; i++)
        {
            scanf("%d %d",&pos[i].x,&pos[i].y);
            hash_insert(pos[i].x, pos[i].y);
        }
        int ans = 0;
        for(i = 0; i < n-1; i++)
        {
            for(j = i+1; j < n; j++)
            {
                int x1 = pos[i].x, y1 = pos[i].y;
                int x2 = pos[j].x, y2 = pos[j].y;
                int add_x = x1-x2,add_y = y1-y2;
                int x3 = x1+add_y;
                int y3 = y1-add_x;
                int x4 = x2+add_y;
                int y4 = y2-add_x;
                if(find(x3,y3) && find(x4,y4))
                    ans++;

                 x3 = x1-add_y;
                 y3 = y1+add_x;
                 x4 = x2-add_y;
                 y4 = y2+add_x;
                if(find(x3,y3) && find(x4,y4))
                    ans++;
            }
        }
        printf("%d\n",ans/4);
    }
    return 0;
}