NYOJ 141 Squares （数学）

描述

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.

输入

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.You can assume the number of test cases is less than 20
输出

For each test case, print on a line the number of squares one can form from the given stars.
样例输入

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
样例输出

1

6

1

分析：

题目的意思其实很简单，就是对于给出的一系列二维坐标系中的点，其中的四个点构成一个正方形，这样不同的正方形一共有多少个。

代码：

#include<stdio.h>

#include<iostream>

#include<algorithm>

using namespace std;

int n;

struct Node

{

    int x;

    int y;

} node[1009];

bool cmp(Node a,Node b)///这些点如果横坐标相同的话，就按照纵坐标从小到大排序；否则就直接按照横坐标从小到大排序

{

    if(a.x==b.x)

        return a.y<b.y;

    else

        return a.x<b.x;

}

bool Search(int x,int y)///采用二分查找法，查一下求出的点是否在已有点的序列中

{

    int left=0;

    int right=n;

    int mid;

    while(left<=right)

    {

        mid=(left+right)/2;

        if(node[mid].x==x&&node[mid].y==y)

            return true;

        else if(node[mid].x==x&&node[mid].y<y||node[mid].x<x)///递归在右区间中找

            left=mid+1;

        else///递归在左区间中找

            right=mid-1;

    }

    return false;

}

int main()

{

    while(~scanf("%d",&n)&&n)

    {

        int ans=0;

        for(int i=0; i<n; i++)

            scanf("%d%d",&node[i].x,&node[i].y);

        sort(node,node+n,cmp);

        int x1,y1,x2,y2;

        for(int i=0; i<n; i++)

            for(int j=i+1; j<n; j++)

            {

                ///这样找的话相当于找的是这条线上方或左方的图形，不理解的话自己用坐标试试

                x1=node[j].x-(node[j].y-node[i].y);

                y1=node[j].y+(node[j].x-node[i].x);

                if(!Search(x1,y1))  continue;

                x2=node[i].x-(node[j].y-node[i].y);

                y2=node[i].y+(node[j].x-node[i].x);

                if(!Search(x2,y2)) continue;

                ans++;

            }

        printf("%d\n",ans/2);///每一个图形都可以由两条边找到，下面和右面的边

    }

    return 0;

}

巴特西

NYOJ 141 Squares （数学）

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