NYOJ 141 Squares （数学）

题目链接

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;
}