POJ 2002 点hash

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 15489		Accepted: 5864

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

二维平面给定一堆点，求能够组成正方形的个数，点数为1000。因此不能枚举四个点推断。

比較优化的方法是将全部点hash。然后枚举两个点，计算出另外两个点的坐标，然后在hash表里查找，最后结果除以4，由于每一条边被统计了4次。

代码：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/11 8:26:00
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=100009;
class HASH{
public:
struct Node{
int next,to;
Node(int _next=0,int _to=0){
next=_next;to=_to;
}
}edge[10010];
int tol,head[maxn+10];
void clear(){
memset(head,-1,sizeof(head));tol=0;
}
void add(int x,int y){
if(find(x,y))return;
int t=(x+maxn)%maxn;
edge[tol]=Node(head[t],y);
head[t]=tol++;
}
int find(int x,int y){
int t=(x+maxn)%maxn;
for(int i=head[t];i!=-1;i=edge[i].next)
if(edge[i].to==y)
return 1;
return 0;
}
}mi;
int x[1010],y[1010];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int n;
while(~scanf("%d",&n)&&n){
mi.clear();
for(int i=0;i<n;i++){
scanf("%d%d",&x[i],&y[i]);
mi.add(x[i],y[i]);
// cout<<"hhh "<<endl;
}
ll ans=0;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++){
//	 cout<<"ddd"<<endl;
ll x3,y3,x4,y4;
x3=x[i]+(y[j]-y[i]);y3=y[i]-(x[j]-x[i]);
x4=x[j]+(y[j]-y[i]);y4=y[j]-(x[j]-x[i]);
if(mi.find(x3,y3)&&mi.find(x4,y4))ans++;
x3=x[i]-(y[j]-y[i]);y3=y[i]+(x[j]-x[i]);
x4=x[j]-(y[j]-y[i]);y4=y[j]+(x[j]-x[i]);
if(mi.find(x3,y3)&&mi.find(x4,y4))ans++;
// cout<<"ddd"<<endl;
}
ans/=4;
cout<<ans<<endl;
}
return 0;
}