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

Sample Output

巴特西

Squares - poj 2002(hash)

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