ACM: 几何题+hash题 poj 2002

Squares

Description

A square is a 4-sided polygonwhose sides have equal length and adjacent sides form 90-degreeangles. It is also a polygon such that rotating about its centre by90 degrees gives the same polygon. It is not the only polygon withthe latter property, however, as a regular octagon also has thisproperty.

So we all know what a square looks like, but can we find allpossible squares that can be formed from a set of stars in a nightsky? To make the problem easier, we will assume that the night skyis a 2-dimensional plane, and each star is specified by its x and ycoordinates.

Input

The input consists of a numberof test cases. Each test case starts with the integer n (1<= n <= 1000) indicating the numberof points to follow. Each of the next n lines specify the x and ycoordinates (two integers) of each point. You may assume that thepoints are distinct and the magnitudes of the coordinates are lessthan 20000. The input is terminated when n = 0.

Output

For each test case, print on aline 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

题意: 现在给你一个坐标内的点集, 要求你判断出有多少个正方形, 四个点不同顺序组成的正方形算一个.

解题思路:
         1. 枚举四个点, 判断是否能组成正方形, 太暴力了, 不可取.
         2. 想法是枚举2个点, 判断另外两个点是否存在. （这样做会出现一个正方行重复4次计算）





     推到另外两个点的坐标：
      很容易得四个直角三角形全等,分别设两条直角边为L1,L2, L1',L2', L1'',L2'', L1''',L2'''
      并且由全等定理有：L1 = L1‘ = L1’‘ = L1’‘’ , L2 = L2' = L2'' = L2''';
      |L1| = y3-y1; |L1'| = x2-x1;
      |L2| = x3-x1; |L2'| = y1-y2;
      得：x3 = x1+(y1-y2); y3 = y1-(x1-x2);
      同理得：x4 = x2+(y1-y2); y4 = y2-(x1-x2);
      另外一边的正方形情况同理可得：
        x3 = x1-(y1-y2); y3 = y1+(x1-x2);
        x4 = x2-(y1-y2); y4 = y2+(x1-x2);

         3. 接下来就是用hash的开散列方法, 快速查找另外2个推到出来的点是否再点集里面.
            我采用的方法是 value = (x*x+y*y) % PRIME;

代码:
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 1005
#define PRIME 99983

struct node
{
    int x, y;
}pos[MAX];

int n;
int hash[PRIME], next[PRIME];

bool find(int x,int y)
{    
    for(int e = hash[(x*x+y*y)%PRIME]; e != -1; e = next[e])
    {
        if(pos[e].x == x && pos[e].y == y)
            return true;
    }
    return false;
}

int main()
{
//    freopen("input.txt","r",stdin);
    while(scanf("%d",&n) != EOF)
    {
        if(n == 0) break;
        
        memset(hash,-1,sizeof(hash));
        for(int i = 0; i < n; ++i)
        {
            scanf("%d %d",&pos[i].x,&pos[i].y);
            int value = pos[i].x*pos[i].x + pos[i].y*pos[i].y;
            value %= PRIME;
            next[i] = hash[value];
            hash[value] = i;
        }
        
        int result = 0;
        for(int i = 0; i < n-1; ++i)
        {
            for(int j = i+1; j < n; ++j)
            {
                int a = pos[i].x - pos[j].x;
                int b = pos[i].y - pos[j].y;
    
                int x3 = pos[i].x + b;
                int y3 = pos[i].y - a;
                int x4 = pos[j].x + b;
                int y4 = pos[j].y - a;
                if(find(x3,y3) && find(x4,y4))
                    result++;
        
                x3 = pos[i].x - b;
                y3 = pos[i].y + a;
                x4 = pos[j].x - b;
                y4 = pos[j].y + a;
                if(find(x3,y3) && find(x4,y4))
                    result++;
            }
        }
        printf("%d\n",result/4);
    }
    return 0;
}