poj2002——Squares（hash）

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

思想是先找出两个点，看另外两个点是否存在。
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)

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <math.h>#include <algorithm>#include <queue>#include <iomanip>#include <map>#define INF 0x3f3f3f3f#define MAXN 10005#define Mod 20007using namespace std;int n,head[20007],next[20007],m;struct Node{    int x,y;};Node star[3000];void add(int i){    int key=(star[i].x*star[i].x+star[i].y*star[i].y)%Mod;    next[m]=head[key];    star[m].x=star[i].x;    star[m].y=star[i].y;    head[key]=m++;}int find(int x,int y){    int key=(x*x+y*y)%Mod;    for(int i=head[key];i!=-1;i=next[i])    {        if(star[i].x==x&&star[i].y==y)            return i;    }    return -1;}bool cmp(Node a,Node b){    if(a.x!=b.x)        return a.x<b.x;    return a.y<b.y;}int main(){    while(~scanf("%d",&n)&&n)    {        memset(head,-1,sizeof(head));        memset(next,0,sizeof(next));        m=1005;        for(int i=0; i<n; ++i)        {            scanf("%d%d",&star[i].x,&star[i].y);            add(i);        }        sort(star,star+n,cmp);        int ans=0;        for(int i=0;i<n;++i)        {            for(int j=i+1;j<n;++j)            {                int x1,y1,x2,y2;                x1=star[i].x-star[j].y+star[i].y;                y1=star[i].y+star[j].x-star[i].x;                if(find(x1,y1)==-1)                    continue;                x2=star[j].x-star[j].y+star[i].y;                y2=star[j].y+star[j].x-star[i].x;                if(find(x2,y2)==-1)                    continue;                ans++;            }        }        printf("%d\n",ans/2);    }    return 0;}