poj 2002 Squares 几何二分 || 哈希

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 15137		Accepted: 5749

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

Source

Rocky Mountain 2004

 

/*
题意：给你1000个点的坐标(x,y)，让你找出能
            构成正方形的个数。
思路：由于是1000，则枚举两个点，求出相应的另外
两个点的坐标。然后用二分判断是否两个点都存在。


就个人而言，关键在  "求出相应的另外两个点的坐标"
设两个点a1,a2;
由a1为中心，逆时针旋转求出
a3.x=a1.y-a2.y+a1.x;
a3.y=a2.x-a1.x+a1.y;
由a2为中心，顺时针旋转求出
a4.x=a1.y-a2.y+a2.x;
a4.y=a2.x-a1.x+a2.y;
由于被计算两次，所以除2
*/
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

typedef struct
{
    int x,y;
}node;
node a[1003];
bool cmp(node n1,node n2)
{
    if( n1.x!=n2.x )
        return n1.x<n2.x;
    else return n1.y<n2.y;
}
bool query(int l,int r,node cur)
{
    int mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if( a[mid].x<cur.x || (a[mid].x==cur.x&&a[mid].y<cur.y))
            l=mid+1;
        else if( a[mid].x>cur.x || ( a[mid].x==cur.x&&a[mid].y>cur.y))
            r=mid-1;
        if( a[mid].x==cur.x && a[mid].y==cur.y) return true;
    }
    return false;
}
int main()
{
    int n,i,j,num;
    node a1,a2,a3,a4;
    while(scanf("%d",&n)>0)
    {
        if(n==0)break;
        for(i=1;i<=n;i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        sort(a+1,a+1+n,cmp);

        for(i=1,num=0;i<n;i++)
        {
            a1=a[i];
            for(j=i+1;j<=n;j++)
            {
                a2=a[j];
                a3.x=a1.y-a2.y+a1.x;
                a3.y=a2.x-a1.x+a1.y;
                if( !query(1,n,a3)) continue;
                a4.x=a1.y-a2.y+a2.x;
                a4.y=a2.x-a1.x+a2.y;
                if( query(1,n,a4)) num++;

            }
        }
        printf("%d
",num/2);
    }
    return 0;
}

哈希做法：

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

const int INF = 1000;
typedef struct
{
    int x,y;
}node;
struct hash
{
    int x;
    int y;
    struct hash *next;
};
struct hash hash_table[1003];
node a[INF+3];

bool cmp(node n1,node n2)
{
    if( n1.x!=n2.x )
        return n1.x<n2.x;
    else return n1.y<n2.y;
}
void Insert(int x,int y)
{
    unsigned k=(x*x+y*y)%INF;
    struct hash *new_hash;
    new_hash=(struct hash *)malloc(sizeof(struct hash));
    new_hash->x=x;
    new_hash->y=y;//build 

    new_hash->next=hash_table[k].next;
    hash_table[k].next=new_hash;
}
bool found(int x,int y)
{
    unsigned k=(x*x+y*y)%INF;
    struct hash *new_hash;
    new_hash=hash_table[k].next;
    while(new_hash!=NULL)
    {
        if(new_hash->x==x && new_hash->y==y)break;
        else new_hash=new_hash->next;
    }
    if(new_hash==NULL)return false;
    else return true;
}

int main()
{
    int n,i,j,num;
    node a1,a2,a3,a4;
    while(scanf("%d",&n)>0)
    {
        if(n==0)break;
        memset(hash_table,0,sizeof(hash_table));
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            Insert(a[i].x,a[i].y);
        }
        sort(a+1,a+1+n,cmp);

        for(i=1,num=0;i<n;i++)
        {
            a1=a[i];
            for(j=i+1;j<=n;j++)
            {
                a2=a[j];
                a3.x=a1.y-a2.y+a1.x;
                a3.y=a2.x-a1.x+a1.y;
                if(!found(a3.x,a3.y))continue;
                a4.x=a1.y-a2.y+a2.x;
                a4.y=a2.x-a1.x+a2.y;
                if(found(a4.x,a4.y)==true)
                    num++;
            }
        }
        printf("%d
",num/2);
    }
    return 0;
}