poj2002 Squares

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 18718 Accepted: 7209

Description

A square is a 4-sided polygon(多边形) 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(八边形) 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

这个题的意思大概就是告诉你N个星星点的坐标，问你这些点最多能组成多少个正方形，4个点的暴力肯定是不行的，肯定是又得想一个暴力两个点的算法，搜了一下题解发现了一个公式：

知道两个点（x1,y1）(x2,y2)那么剩余的两个点

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)

利用三角形全等可以证的，这两个点就是一条边上的两个点，然后分别向两侧（以这条边为斜边）做直角三角形，然后分别连接剩余的点，同样以连接的边做直角三角形，两个三角形是全等的，便可以得到坐标关系。

有兴趣的可以自己试一下(≧▽≦)

这样我们可以先把每个点的坐标记录下来，然后暴力两个点，看看剩余的两个点是不是存在即可。

由于这种暴力每条边都会搜索，所以我们每个正方形会记录4次，最后累加和的结果要除以4.当然直接搜索跟暴力4个点一样，我们用哈希表散列一下，

这里我直接是哈希表的下标=(x^2+y^2)%1007,试过几次，发现这个是最快的，这个空间利用率比较高是毋庸置疑的，表越大，搜索起来会更快，当然是跟数据是有关的。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <vector>#include <map>using namespace std;const int MAXN=1007;const int inf=0x3f3f3f;struct node{    int x,y;};int x[1005],y[1005];vector<node>head[MAXN];bool findx(int x,int y){    int d=(x*x+y*y)%MAXN,i;    int l=head[d].size();    for(i=0;i<l;++i)    {        if(head[d][i].x==x&&head[d][i].y==y)return 1;    }    return 0;}int main(){    int i,j;    int n;    while(~scanf("%d",&n))    {        if(!n)break;        node t;        int ha;        for(i=0;i<MAXN;++i)head[i].clear();        for(i=0;i<n;++i)        {            scanf("%d%d",&x[i],&y[i]);            t.x=x[i];t.y=y[i];            ha=(t.x*t.x+t.y*t.y)%MAXN;            head[ha].push_back(t);        }        int ans=0;        int lx,ly,x1,y1,x2,y2;        for(i=0;i<n-1;++i)            for(j=i+1;j<n;++j)        {             lx=x[i]-x[j];             ly=y[i]-y[j];             x1=x[i]+ly,y1=y[i]-lx;             x2=x[j]+ly,y2=y[j]-lx;            if(findx(x1,y1)&&findx(x2,y2))ans++;            x1=x[i]-ly;y1=y[i]+lx;            x2=x[j]-ly;y2=y[j]+lx;            if(findx(x1,y1)&&findx(x2,y2))ans++;        }        printf("%d\n",ans>>2);    }    return 0;}

http://www.cnblogs.com/luyingfeng/archive/2013/08/23/3277094.html

公式是从这里看的，但是发现这人完全是水过去的，暴力的姿势非常正确导致完全没问题(╯‵□′)╯︵┻━┻我也是惊呆了，表示他完全没看坐标的范围啊喂