Squares 分类： POJ 2015-08-04 11:46 3人阅读 评论(0) 收藏

Squares

Time Limit: 3500MS Memory Limit: 65536K

Total Submissions: 17462 Accepted: 6634

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

在推正方形顶点时,多亏了金巨巨,金巨巨就是给力

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)

const int MAX = 1010;

struct node
{
    int x;
    int y ;
}Point[MAX];

bool cmp(node b,node c)
{
    if(b.x<c.x||(b.x==c.x&&b.y<c.y))
    {
        return true;
    }
    return false;
}

bool Look(int low,int high,int x,int y)//二分查找
{
    int i=low,j=high;
    while(i<=j)
    {
        int mid=(i+j)/2;
        if(Point[mid].x==x&&Point[mid].y==y)
        {
            return true;
        }
        if(Point[mid].x<x)
        {
            i=mid+1;
        }
        else if(Point[mid].x>x)
        {
            j=mid-1;
        }
        else if(Point[mid].x==x)
        {
            if(Point[mid].y<y)
            {
                i=mid+1;
            }
            else if(Point[mid].y>y)
            {
                j=mid-1;
            }
        }
    }
    return false;
}
int main()
{

    int n;
    while(scanf("%d",&n),n)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&Point[i].x,&Point[i].y);
        }
        sort(Point,Point+n,cmp);
        int sum=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                int ans=(Point[i].x+Point[j].x-Point[j].y+Point[i].y);//计算正方形的其余的两个顶点
                int ant=(Point[i].y+Point[j].y+Point[j].x-Point[i].x);
                bool flag=false;
                if(ans%2==0&&ant%2==0)
                {
                    flag=Look(0,n-1,ans/2,ant/2);
                }

                if(!flag)
                {
                    continue;
                }
                flag=false;
                ans=(Point[i].x+Point[j].x+Point[j].y-Point[i].y);
                ant=(Point[i].y+Point[j].y-Point[j].x+Point[i].x);
                if(ans%2==0&&ant%2==0)
                {
                    flag=Look(0,n-1,ans/2,ant/2);
                }
                if(flag)
                {
                    sum++;
                }
            }
        }
        printf("%d\n",sum/2);//对于每个正方形都会查找到两次,所以除二
    }
    return 0;
}

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