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[];
 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)/;
         if( a[mid].x<cur.x || (a[mid].x==cur.x&&a[mid].y<cur.y))
             l=mid+;
         else if( a[mid].x>cur.x || ( a[mid].x==cur.x&&a[mid].y>cur.y))
             r=mid-;
         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)>)
     {
         if(n==)break;
         for(i=;i<=n;i++)
             scanf("%d%d",&a[i].x,&a[i].y);
         sort(a+,a++n,cmp);

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

             }
         }
         printf("%d\n",num/);
     }
     return ;
 }

哈希做法：

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

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

 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)>)
     {
         if(n==)break;
         memset(hash_table,,sizeof(hash_table));
         for(i=;i<=n;i++)
         {
             scanf("%d%d",&a[i].x,&a[i].y);
             Insert(a[i].x,a[i].y);
         }
         sort(a+,a++n,cmp);

         for(i=,num=;i<n;i++)
         {
             a1=a[i];
             for(j=i+;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\n",num/);
     }
     return ;
 }