POJ 2002 Squares

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 18718 Accepted: 7209

Description

A square is a 4-sided polygon(多边形) whose sides have equal length and adjacent(邻近的) sides form 90-degree angles. It is also a polygon such that rotating(旋转的) 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(八边形) 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(承担) that the night sky is a 2-dimensional plane, and each star is specified(指定) by its x and y coordinates(坐标). 

Input

The input(投入) consists of a number of test cases. Each test case starts with the integer(整数) n (1 <= n <= 1000) indicating(表明) 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(明显的) and the magnitudes(大小) of the coordinates are less than 20000. The input is terminated(终止) 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

因为坐标有负数，所以在生成key值时要注意一下，思路很简单，先枚举两个点所有的组合，然后通过数学公式的换算来计算另外两个点，然后用哈希表查找一下那两个点是否存在。因为我们把所有点都枚举过一次，所以一个正方形被我们找过四次，把最后结果除以四就好了。

#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<vector>#include<map>#include<ctime>#include<cstdlib>#include<algorithm>#include<new>using namespace std;const int mod = 100007;struct node{    int x;    int y;}dian[1234];node *hs[mod+6];bool f(int xx,int yy,int xxx,int yyy){    long long int t;    int flag=0;    t=xx*xx+yy*yy;    t=t%mod+1;    while(hs[t])    {        if(hs[t]->x==xx&&hs[t]->y==yy)        {            flag++;            break;        }        t++;    }     t=xxx*xxx+yyy*yyy;    t=t%mod+1;    while(hs[t])    {        if(hs[t]->x==xxx&&hs[t]->y==yyy)        {            flag++;            break;        }        t++;    }    if(flag==2) return 1;    else return 0;}int main(){    int n;    while(cin>>n&&n)    {        memset(hs,0,sizeof(hs));        for(int i=1;i<=n;i++)        {            scanf("%d%d",&dian[i].x,&dian[i].y);            long long int key=(dian[i].x*dian[i].x)+(dian[i].y*dian[i].y);            key=key%mod+1;            if(hs[key]) while(hs[key]) ++key;          //  hs[key]=new node;            hs[key]=&dian[i];        }        int sum=0;        for(int i=2;i<=n;i++)        for(int j=1;j<i;j++)        {            int x1=dian[i].x,x2=dian[j].x,y1=dian[i].y,y2=dian[j].y,x3,x4,y3,y4;            x3=x1+(y1-y2);            x4=x2+(y1-y2);            y3=y1-(x1-x2);            y4=y2-(x1-x2);            if(f(x3,y3,x4,y4)) sum++;            x3=x1-(y1-y2);            x4=x2-(y1-y2);            y3=y1+(x1-x2);            y4=y2+(x1-x2);            if(f(x3,y3,x4,y4)) sum++;        }        cout<<sum/4<<endl;    }    return 0;}