POJ2002Squares（哈希）

http://poj.org/problem?id=2002

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

题目大意就是给一些点的坐标，问最多可以构成多少个正方形。

直接四个点四个点地枚举肯定超时的，不可取。

普遍的做法是：先枚举两个点，通过数学公式得到另外2个点，使得这四个点能够成正方形。然后检查散点集中是否存在计算出来的那两个点，若存在，说明有一个正方形。

但这种做法会使同一个正方形按照不同的顺序被枚举了四次，因此最后的结果要除以4.

 

已知： (x1,y1)  (x2,y2)

则：   x3=x1+(y1-y2)   y3= y1-(x1-x2)

x4=x2+(y1-y2)   y4= y2-(x1-x2)

或

x3=x1-(y1-y2)   y3= y1+(x1-x2)

x4=x2-(y1-y2)   y4= y2+(x1-x2)

 

据说是利用全等三角形可以求得上面的公式

有兴趣的同学可以证明下。。。

 

这样就好做了，用哈希标记每一个点这样就是O(n^2)的算法了，n的范围是1000这肯定不会超时

具体建议参考ζёСяêτ - 小優YoU的博客http://blog.csdn.net/lyy289065406/article/details/6647405,说的相当清楚

 

下面附代码

#include <iostream>
#include <cstdio>
#include<cstring>
using namespace std;

const int prime = 100007;
int hash[prime+5];
int next[prime+5];

struct node
{
    int x,y;
}nod[1005];

void insert(int x,int y,int index)//加入哈希链表中
{
    int h = (x*x+y*y) % prime;//哈希函数就是平方和取余
    int u = hash[h];
    while(u)
    {
        u=next[u];
    }
    next[index]=hash[h];
    hash[h]=index;
}

bool find(int x,int y)
{
    int h = (x*x+y*y) % prime;
    int u = hash[h];
    while(u)
    {
        if(nod[u].x==x && nod[u].y==y)return true;
        u=next[u];
    }
    return false;
}

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        memset(hash,0,sizeof(hash));
        memset(next,0,sizeof(next));

        int i,j;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&nod[i].x,&nod[i].y);
            insert(nod[i].x,nod[i].y,i);
        }

        int x1,y1,x2,y2,count=0;
        for(i=1;i<n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                x1=nod[i].x+(nod[i].y-nod[j].y);
                y1=nod[i].y-(nod[i].x-nod[j].x);
                x2=nod[j].x+(nod[i].y-nod[j].y);
                y2=nod[j].y-(nod[i].x-nod[j].x);
                if(find(x1,y1)&&find(x2,y2))count++;

                x1=nod[i].x-(nod[i].y-nod[j].y);
                y1=nod[i].y+(nod[i].x-nod[j].x);
                x2=nod[j].x-(nod[i].y-nod[j].y);
                y2=nod[j].y+(nod[i].x-nod[j].x);
                if(find(x1,y1) && find(x2,y2))count++;
            }
        }
        printf("%d\n",count/4);
    }
    return 0;
}