poj2002 数正方形 （哈希+几何）

题目传送门

题目大意：给你一堆点，问你能组成几个正方形。

思路：一开始想的是用对角线的长度来当哈希的key，但判断正方形会太复杂，然后就去找了一下正方形的判断方法，发现

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)

就是枚举两个点，然后算出另外两个点。在哈希表中看看能不能找到这两个点。如果只采取其中一个公式的话，切记点的坐标要排序再哈希（具体原因我也不知道，但自己举了很多样例确实是这样），你也可以把两个公式都用上，两个if得到答案。

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const double PI=acos(-1.0);
int fact[10]= {1,1,2,6,24,120,720,5040,40320,362880};
const int maxn= 1010;
const int mod = 100019;
int n;
struct dian {
	int x,y;
} a[maxn];
int hash[mod+10];
int next[maxn];
int gethash(int i) {
	return (a[i].x*a[i].x%mod+a[i].y*a[i].y%mod)%mod;
}
bool find(int x,int y){
	int hashval=(x*x%mod+y*y%mod)%mod;
	for(int i=hash[hashval];i!=-1;i=next[i])
	{
		if(a[i].x==x&&a[i].y==y)return true;
	}
	return false;
}
bool cmp(dian aa,dian bb){
	if(aa.x!=bb.x)
	return aa.x<bb.x;
	return aa.y<bb.y;
}
int main(){
	while(scanf("%d",&n),n)
	{
		int ans=0;
		memset(hash,-1,sizeof(hash));
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&a[i].x,&a[i].y);
		}
		sort(a+1,a+1+n,cmp);//非常重要  没有则wa
		for(int i=1;i<=n;i++)
		{
			int hashval=gethash(i);//映射到哈希表里
			next[i]=hash[hashval];
			hash[hashval]=i;
		}
		for(int i=1;i<n;i++)
		{
			for(int j=i+1;j<=n;j++)
			{
				int x=a[i].x-a[j].y+a[i].y;//算出第一个点
				int y=a[i].y+a[j].x-a[i].x;
				if(!find(x,y))continue;//查找
				x=a[j].x-a[j].y+a[i].y;//第二个点
				y=a[j].y+a[j].x-a[i].x;
				if(!find(x,y))continue;//查找
				ans++;
			}
		}
	printf("%d\n",ans/2);//由于有重复计算  所以要除以二
	}
}

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 21208		Accepted: 8136

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