Squares-暴力枚举或者二分

B - Squares

Time Limit:3500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2002

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

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

/*Author: 2486Memory: 24256 KBTime: 375 MSLanguage: C++Result: Accepted*///此题目暴力暴力枚举//通过已经确定好的两点，算出剩下的两点//（有两种情况）//一个在上面，一个下面#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=20000+5;struct point{    int x,y;}ps[1005];int n,ans;bool vis[maxn<<1][maxn<<1];int main(){while(~scanf("%d",&n),n){    ans=0;    for(int i=0;i<n;i++){        scanf("%d%d",&ps[i].x,&ps[i].y);        ps[i].x+=20000,ps[i].y+=20000;//在数组里面可以存储负数        vis[ps[i].x][ps[i].y]=true;//标记着这个点存在    }    for(int i=0;i<n;i++){        for(int j=0;j<i;j++){            if(i==j)continue;//分别代表着上下两种不同的正方形            int nx1=ps[i].x+ps[i].y-ps[j].y;            int ny1=ps[i].y+ps[j].x-ps[i].x;            int nx2=ps[j].x+ps[i].y-ps[j].y;            int ny2=ps[j].y+ps[j].x-ps[i].x;            if(vis[nx1][ny1]&&vis[nx2][ny2])ans++;            nx1=ps[i].x-(ps[i].y-ps[j].y);            ny1=ps[i].y-(ps[j].x-ps[i].x);            nx2=ps[j].x-(ps[i].y-ps[j].y);            ny2=ps[j].y-(ps[j].x-ps[i].x);            if(vis[nx1][ny1]&&vis[nx2][ny2])ans++;        }    }    for(int i=0;i<n;i++){        vis[ps[i].x][ps[i].y]=false;//必须要进行清零，不能用memset，因为数组有点大    }    printf("%d\n",ans/4);}return 0;}