POJ 2202 哈希

Squares
Time Limit: 3500MS Memory Limit: 65536K
Total Submissions: 18116 Accepted: 6946
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

题意：
给出坐标轴上的多个点，问最多这些点可以组成多少个正方形。
题解：
直接暴力枚举肯定是不行的。考虑只要知道了两个点，就能推出有可能的两个正方形。然后做一个hash表，查找这两个点是否在哈希表里。最后可知一个正方形四个边。最后的结果需要除以4。用数学推点的时候需要用到高中的数学知识。推导过程不算特别麻烦，这里直接给出结论。
已知两个点A（x1,y1）,B(x2,y2).
则其中一个正方形的两点为：
C1（x2 + y1 - y2,y2 + x2 - x1）
D1（x1 + y1 - y2,y1 + x2 - x1）
另一个正方形的两点为：
C2（x2 + y2 - y1,y2 + x1 - x2）
D2（x1 + y2 - y1,y1 + x1 - x2）
hash键值k = （x^2+y^2）%M

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cstdlib>#define M 10007using namespace std;struct Po{    int x,y;    Po* next;};Po G[M];Po* Hash[M+5] = {NULL};void init(void){    for(int i = 0;i<M+5;i++)        Hash[i] = NULL;}bool IsFind(Po tem){    int k = (tem.x*tem.x+tem.y*tem.y)%M;    if(Hash[k] == NULL)        return false;    else{        Po* p = Hash[k];        while(p!=NULL){            if(tem.x==p->x&&tem.y==p->y)                return true;            p = p->next;        }    }    return false;}int main(){//    freopen("data.txt","r",stdin);    int n;    while(scanf("%d",&n)&&n)    {        init();        int i,j;        for(i = 0;i<n;i++){            int a,b;            scanf("%d%d",&a,&b);            G[i].x = a;            G[i].y = b;            int k = (a*a+b*b)%M;            Po* tem = (Po*)malloc(sizeof(Po));            tem->next = NULL;            tem->x = a;            tem->y = b;            if(Hash[k]==NULL) Hash[k] = tem;            else            {                tem->next = Hash[k];                Hash[k] = tem;            }        }        long long sum = 0;        for(i = 1;i<n;i++)            for(j = 0;j<i;j++){                if(i!=j){                    int x1 = G[i].x;                    int y1 = G[i].y;                    int x2 = G[j].x;                    int y2 = G[j].y;                    Po tem;                    int ok1 = 0;                    tem.x = x2+y1-y2;                    tem.y = y2+x2-x1;                    if(IsFind(tem)){                        tem.x = x1+y1-y2;                        tem.y = y1+x2-x1;                        if(IsFind(tem))                            ok1 = 1;                    }                    int ok2 = 0;                    tem.x = x2+y2-y1;                    tem.y = y2+x1-x2;                    if(IsFind(tem)){                        tem.x = x1+y2-y1;                        tem.y = y1+x1-x2;                        if(IsFind(tem))                            ok2 = 1;                    }                    if(ok1) sum++;                    if(ok2) sum++;                }            }        printf("%lld\n",sum/4);    }    return 0;}