POJ 2002-Squares（hash拉链法-正方形个数）

红太狼 2022-09-29 05:21 112阅读 0赞

Squares

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;3500MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;19942</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;7674</td> 
  </tr> 
 </tbody> 
</table>

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][]

## 题目意思： ##

给出N个点的坐标，求用其中4个点能组成的正方形的个数，顺序不同计为重复。

## 解题思路： ##

hash表使用拉链法解决冲突。

依次枚举两个点，根据这两个点计算能够构成正方形的其他两个点的坐标。

![20170314200451058][]

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<map>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define MAXN 100007
    #define INF 0x3f3f3f3f
    struct se
    {
        int x,y;
    } s[1010];
    int ha[MAXN+5];//ha[i]表示键值i的下标
    int next[MAXN+5];//next[i]表示对于当前位置下标i，如果ha中的键值重复，存储上一个该键值的下标；否则为0表示无重复
    
    void Insert(int x,int y,int pos)//拉链法解决冲突
    {
        int t=(x*x+y*y)%MAXN,q=ha[t];
        while(q)//已经存在
            q=next[q];
        next[pos]=ha[t];
        ha[t]=pos;
    }
    
    bool Find(int x,int y)
    {
        int t=(x*x+y*y)%MAXN,q=ha[t];
        while(q)//存在
        {
            if(s[q].x==x&&s[q].y==y) return true;
            q=next[q];
        }
        return false;
    }
    
    int main()
    {
    
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        int n;
        while(cin>>n&&n)
        {
            int ans=0;
            memset(ha,0,sizeof(ha));
            memset(next,0,sizeof(next));
            for(int i=1; i<=n; ++i)
            {
                cin>>s[i].x>>s[i].y;
                Insert(s[i].x,s[i].y,i);
            }
            for(int i=1; i<=n; ++i)
                for(int j=i+1; j<=n; ++j)
                {
                    se a,b;
                    //另外两个点在上方
                    a.x=s[i].x-(s[j].y-s[i].y);
                    a.y=s[i].y+(s[j].x-s[i].x);
                    b.x=a.x+(s[j].x-s[i].x);
                    b.y=a.y+(s[j].y-s[i].y);
                    if(Find(a.x,a.y)&&Find(b.x,b.y)) ++ans;
                    //两外两个点在下方
                    a.x=s[i].x+(s[j].y-s[i].y);
                    a.y=s[i].y-(s[j].x-s[i].x);
                    b.x=a.x+(s[j].x-s[i].x);
                    b.y=a.y+(s[j].y-s[i].y);
                    if(Find(a.x,a.y)&&Find(b.x,b.y)) ++ans;
                }
            cout<<ans/4<<endl;
        }
        return 0;
    }
    /*
    1 1
    1
    5 3
    4 5 3 2 1
    */

[Rocky Mountain 2004]: http://poj.org/searchproblem?field=source&key=Rocky+Mountain+2004
[20170314200451058]: /images/20220708/a5569741a4494a6bba27dfbce1ec582c.png