poj2002 Squares(hash+折半枚举)

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：在二维平面上，给你n个点的坐标，求任取四点构成正方形的个数
题解：好吧，暴力O(n^4)又t了……
那么折半枚举吧
枚举两个点连成一条线，然后根据这条线可以得到正方形的另外两个点，根据hash可以快速确定点是否在n个点中，即可以计算出正方形的个数
代码如下：

#include<map>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;

struct node
{
    int x,y;
} a[1010];

vector<long long> g[100010];
int n;

int cmp(node x,node y)
{
    if(x.x==y.x)
    {
        return x.y<y.y;
    }
    return x.x<y.x;
}

int main()
{

    while(scanf("%d",&n)==1&&n)
    {
        int ans=0;
        vector<long long> g[100010];
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
        }
        sort(a+1,a+n+1,cmp);
        for(int i=1; i<=n; i++)
        {
            long long key=(a[i].x*a[i].x+a[i].y*a[i].y)%99991;
            g[key].push_back(i);
        }
        for(int i=1; i<=n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                int x1,y1,x2,y2;
                int dx=(a[j].x-a[i].x),dy=(a[j].y-a[i].y);
//            if(dx<0||dy<0)
//            {
//                continue;
//            }
                x1=a[i].x-dy;
                y1=a[i].y+dx;
                x2=a[j].x-dy;
                y2=a[j].y+dx;
                int flag=0;
                long long key=(x1*x1+y1*y1)%99991;
                for(int k=0; k<g[key].size(); k++)
                {
                    if(a[g[key][k]].x==x1&&a[g[key][k]].y==y1)
                    {
                        flag+=1;
                    }
                }
                key=(x2*x2+y2*y2)%99991;
                for(int k=0; k<g[key].size(); k++)
                {
                    if(a[g[key][k]].x==x2&&a[g[key][k]].y==y2)
                    {
                        flag+=1;
                    }
                }
                if(flag==2)
                {
//                if(a[i].x==x1&&a[i].y==y1||a[j].x==x2&&a[j].y==y2||a[i].x==x2&&a[i].y==y2||a[j].x==x1&&a[j].y==y1)
//                {
//                    continue;
//                }
//                printf("%d %d\n",dx,dy);
//                printf("%d %d %d %d %d %d %d %d\n",a[i].x,a[i].y,a[j].x,a[j].y,x1,y1,x2,y2);
                    ans++;
                }
            }
        }
        printf("%d\n",ans/2);
    }

}