poj2002 -- 点的hash

题目：

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 15261		Accepted: 5792

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

Rocky Mountain 2004

 

 

大意：给出n个点的坐标  求能组成多少个正方形。

 

思路：

依次枚举两个点，求出另外点的坐标，看看另外能组成正方形的点是否存在。

 

已知两点A（x1，y1）B（x2，y2）

则  另外两点的坐标：

x3=x1+|y1-y2|;

y3=y1-|x1-x2|;

x4=x2+|y1-y2|;

y2=y2-|x1-x2|;

 

最难的部分便是对点的储存了 ,

 

这里先给出 点储存的模板，通过hash数组和next数组的转换  达到储存点的目的：

 

code：

struct point
{
    int x;
    int y;
}poin[maxn];

int hash[maxn+10];
int next[maxn+10];

int findkey(point p)
{
    return abs(p.x+p.y)%maxn;
}

void hashinsert(int i)
{
    int key=findkey(poin[i]);
    next[i]=hash[key];//最后一个next储存的是-1，其他的都指向下一个next
    hash[key]=i;//指向最开头的next的下标
}

int hashsearch(point p)
{
    int key=findkey(p);
    int i=hash[key];
    while(i!=-1)
    {
        if(p.x==poin[i].x&&p.y==poin[i].y)//找到一个这样的点
        return 1;
        i=next[i];
    }
    return 0;
}

  题目代码：

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;

const int maxn=1031;

struct point
{
    int x;
    int y;
}poin[maxn];

int hash[maxn+10];
int next[maxn+10];

int findkey(point p)
{
    return abs(p.x+p.y)%maxn;
}

void hashinsert(int i)
{
    int key=findkey(poin[i]);
    next[i]=hash[key];
    hash[key]=i;
}

int hashsearch(point p)
{
    int key=findkey(p);
    int i=hash[key];
    while(i!=-1)
    {
        if(p.x==poin[i].x&&p.y==poin[i].y)
        return 1;
        i=next[i];
    }
    return 0;
}


bool cmp(point p1,point p2)
{
    if(p1.x!=p2.x)
    return p1.x<p2.x;
    else
    return p1.y<p2.y;
}

int main()
{
    int n;
    int x1,x2,x3,x4;
    int y1,y2,y3,y4;
    //freopen("aa.txt","r",stdin);
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        break;
        memset(hash,-1,sizeof(hash));
        memset(next,-1,sizeof(next));
        point p3,p4;
        for(int i=0;i<n;i++)
            scanf("%d%d",&poin[i].x,&poin[i].y);
        sort(poin,poin+n,cmp);
        for(int i=0;i<n;i++)
            hashinsert(i);
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                x1=poin[i].x;
                y1=poin[i].y;
                x2=poin[j].x;
                y2=poin[j].y;
                p3.x=x1+(y2-y1);
                p3.y=y1-(x2-x1);
                p4.x=x2+(y2-y1);
                p4.y=y2-(x2-x1);
                if(hashsearch(p3)&&hashsearch(p4))
                ans++;
            }
        }
        printf("%d
",ans/2);
    }
    return 0;
}