Squares - poj 2002(hash)

枚举两个点作为一条边，求出正方形的另外两个点,利用hash查找另外两个点。

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 struct node{
     int x,y;
 } point[];

 struct hash{
     int pos;
     hash* next;
 } hashtable[];
 int find(int x,int y){

     hash* t;
     int tmp=(x*x+y*y)%;
     t=&hashtable[tmp];
     if(t->pos==){
         return ;
     }else{
         int pos=t->pos;
         if(point[pos].x==x&&point[pos].y==y)
             return ;
     }
     while(t->next){
         t=t->next;
         int pos=t->pos;
         if(point[pos].x==x&&point[pos].y==y)
             return ;
     }
     return ;
 }
 int N;
 int main() {
     scanf("%d",&N);
     int i,j;
     while(N){
         memset(hashtable,,sizeof(hashtable));
         memset(point,,sizeof(point));
         for(i=;i<=N;i++){
             int x,y;
             scanf("%d %d",&x,&y);
             point[i].x=x;
             point[i].y=y;
             int t=(x*x+y*y)%;
             if(hashtable[t].pos==){
                 hashtable[t].pos=i;
             }else{
                 hash* tmp=(hash*)malloc(sizeof(hash));
                 tmp->next=;
                 tmp->pos=i;
                 tmp->next=hashtable[t].next;
                 hashtable[t].next=tmp;
             }
         }
         int result=;
         for(i=;i<=N;i++){
             for(j=i+;j<=N;j++){
                 int x1=point[i].x+(point[j].y-point[i].y);
                 int y1=point[i].y-(point[j].x-point[i].x);
                 int x2=point[j].x-(point[i].y-point[j].y);
                 int y2=point[j].y+(point[i].x-point[j].x);
                 if(find(x1,y1)&&find(x2,y2))
                     result++;
             }
         }
         for(i=;i<=N;i++){
             for(j=i+;j<=N;j++){
                 int x1=point[i].x-(point[j].y-point[i].y);
                 int y1=point[i].y+(point[j].x-point[i].x);
                 int x2=point[j].x+(point[i].y-point[j].y);
                 int y2=point[j].y-(point[i].x-point[j].x);
                 if(find(x1,y1)&&find(x2,y2))
                     result++;
             }
         }
         printf("%d\n",result/);
         scanf("%d",&N);
     }

     return ;
 }

 

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 17553		Accepted: 6677

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