Squares

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[1002];

struct hash{
    int pos;
    hash* next;
} hashtable[2017];
int find(int x,int y){

    hash* t;
    int tmp=(x*x+y*y)%2017;
    t=&hashtable[tmp];
    if(t->pos==0){
        return 0;
    }else{
        int pos=t->pos;
        if(point[pos].x==x&&point[pos].y==y)
            return 1;
    }
    while(t->next){
        t=t->next;
        int pos=t->pos;
        if(point[pos].x==x&&point[pos].y==y)
            return 1;
    }
    return 0;
}
int N;
int main() {
    scanf("%d",&N);
    int i,j;
    while(N){
        memset(hashtable,0,sizeof(hashtable));
        memset(point,0,sizeof(point));
        for(i=1;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)%2017;
            if(hashtable[t].pos==0){
                hashtable[t].pos=i;
            }else{
                hash* tmp=(hash*)malloc(sizeof(hash));
                tmp->next=0;
                tmp->pos=i;
                tmp->next=hashtable[t].next;
                hashtable[t].next=tmp;
            }
        }
        int result=0;
        for(i=1;i<=N;i++){
            for(j=i+1;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=1;i<=N;i++){
            for(j=i+1;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
",result/4);
        scanf("%d",&N);
    }

    return 0;
}

 

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