poj 2002 Squares

题目链接：http://poj.org/problem?id=2002

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：二维平面给出一些点，计算以这些点为顶点的正方形的个数。

解法：哈希。因为n为1000，坐标大小高达20000，普通查找肯定不行的，我们得用哈希压缩坐标。

　　   然后枚举正方形的一条边上的两个顶点，可以计算出剩余的两个顶点坐标，然后哈希表查找是否有这样的正方形，统计个数，然后除以4即可（因为枚举了每个正方形的每条边）。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
typedef long long ll;
const int maxn=1000+10;
const int mod=100007;
struct node
{
    int x,y;
    int next;
}edge[maxn];
int head[mod],edgenum;
int xx[maxn],yy[maxn];
void insert_hash(int x,int y)
{
    int k=(x*x+y*y)%mod;
    edge[edgenum].x=x ;
    edge[edgenum].y=y ;
    edge[edgenum].next=head[k] ;
    head[k]=edgenum++ ;
}
int search_hash(int x,int y)
{
    int k=(x*x+y*y)%mod;
    for (int i=head[k] ;i!=-1 ;i=edge[i].next)
    {
        if (x==edge[i].x && y==edge[i].y) return true;
    }
    return false;
}
int main()
{
    int n;
    while (scanf("%d",&n)!=EOF && n)
    {
        memset(head,-1,sizeof(head));
        edgenum=0;
        for (int i=0 ;i<n ;i++)
        {
            scanf("%d%d",&xx[i],&yy[i]);
            insert_hash(xx[i],yy[i]);
        }
        int count=0;
        for (int i=0 ;i<n ;i++)
        {
            for (int j=i+1 ;j<n ;j++)
            {
                int x=xx[i]+(yy[j]-yy[i]);
                int y=yy[i]-(xx[j]-xx[i]);
                int x2=xx[j]-(yy[i]-yy[j]);
                int y2=yy[j]+(xx[i]-xx[j]);
                if (search_hash(x,y) && search_hash(x2,y2)) count ++ ;
            }
        }
        for (int i=0 ;i<n ;i++)
        {
            for (int j=i+1 ;j<n ;j++)
            {
                int x=xx[i]-(yy[j]-yy[i]);
                int y=yy[i]+(xx[j]-xx[i]);
                int x2=xx[j]+(yy[i]-yy[j]);
                int y2=yy[j]-(xx[i]-xx[j]);
                if (search_hash(x,y) && search_hash(x2,y2)) count ++ ;
            }
        }
        printf("%d
",count/4);
    }
    return 0;
}

后续：感谢大牛提出宝贵的意见。。。