POJ 2002 点的hash

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal 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

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

二维平面给定一堆点，求可以组成正方形的个数，点数为1000，因此不能枚举四个点判断，

比较优化的方法是将所有点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，因为每一条边被统计了4次。

代码：

/* ***********************************************Author :_rabbitCreated Time :2014/5/11 8:26:00File 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;}