poj 2002(好题 链式hash+已知正方形两点求另外两点)

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 18493		Accepted: 7124

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

题意：平面上有n个点,问n个点能够组成多少个正方形.
初看很简单,就是枚举每两个点去找另外两个点..然后我马上遇到了难题,我准备用bool去存点，结果试了一下 40000*40000直接爆掉了..然后正方形的其余两个点不会求..因为这题都是整形，用斜率求肯定
会出问题（平时做几何体，最好不要用斜率，因为会有正无穷）。然后竟然有公式！！用全等三角形可以证明。。。我等膜拜。。然后就是不能用bool只能用hash了。。hash又是一个好的借鉴。。链式前向星
存的。。

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;
const int maxn = 1005;
const int H = 10000;
struct Point
{
    int x,y;
} p[maxn],C,D;
///************************hash模板

int Hash[H],cur;
void initHash(){
    memset(Hash,-1,sizeof(Hash));
    cur =0;
}
struct Node
{
    int x,y;
    int next;
} node[maxn];
void InsertHash(int x,int y)
{
    int h=(x*x + y*y) % H;
    node[cur].x=x;
    node[cur].y=y;
    node[cur].next=Hash[h];
    Hash[h]=cur++;
}
bool SearchHash(int x,int y)
{
    int h=(x*x + y*y) % H;
    int next=Hash[h];
    while(next != -1)
    {
        if(node[next].x == x && node[next].y == y) return true;
        next = node[next].next;
    }
    return false;
}
/*******************************/
///已知正方形 a,b 两点求解另外两点的 c,d的坐标(要分两种情况)
void solve1(Point a,Point b,Point& c,Point& d)  ///情况1
{
    c.x = a.x + (a.y-b.y);
    c.y = a.y - (a.x-b.x);
    d.x = b.x + (a.y-b.y);
    d.y = b.y - (a.x-b.x);
}
void solve2(Point a,Point b,Point& c,Point& d)  ///情况2(反过来)
{
    c.x = a.x - (a.y-b.y);
    c.y = a.y + (a.x-b.x);
    d.x = b.x - (a.y-b.y);
    d.y = b.y + (a.x-b.x);
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n)
    {
        initHash();
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            InsertHash(p[i].x,p[i].y);
        }
        int ans = 0;
        for(int i=0; i<n; i++)
        {
            for(int j=i+1; j<n; j++)
            {
                solve1(p[i],p[j],C,D);
                if(SearchHash(C.x,C.y)&&SearchHash(D.x,D.y)) ans++;
                solve2(p[i],p[j],C,D);
                if(SearchHash(C.x,C.y)&&SearchHash(D.x,D.y)) ans++;
            }
        }
        printf("%d
",ans/4);
    }
    return 0;
}