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

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

Source

 

1 /* 2 题意：给你1000个点的坐标(x,y)，让你找出能 3             构成正方形的个数。 4 思路：由于是1000，则枚举两个点，求出相应的另外 5 两个点的坐标。然后用二分判断是否两个点都存在。 6  7  8 就个人而言，关键在  "求出相应的另外两个点的坐标" 9 设两个点a1,a2;10 由a1为中心，逆时针旋转求出11 a3.x=a1.y-a2.y+a1.x;12 a3.y=a2.x-a1.x+a1.y;13 由a2为中心，顺时针旋转求出14 a4.x=a1.y-a2.y+a2.x;15 a4.y=a2.x-a1.x+a2.y;16 由于被计算两次，所以除217 */18 #include
      
       19 #include
       
        20 #include
        
         21 #include
         
          22 #include
          
           23 using namespace std;24 25 typedef struct26 {27     int x,y;28 }node;29 node a[1003];30 bool cmp(node n1,node n2)31 {32     if( n1.x!=n2.x )33         return n1.x
           
            cur.x || ( a[mid].x==cur.x&&a[mid].y>cur.y))45 r=mid-1;46 if( a[mid].x==cur.x && a[mid].y==cur.y) return true;47 }48 return false;49 }50 int main()51 {52 int n,i,j,num;53 node a1,a2,a3,a4;54 while(scanf("%d",&n)>0)55 {56 if(n==0)break;57 for(i=1;i<=n;i++)58 scanf("%d%d",&a[i].x,&a[i].y);59 sort(a+1,a+1+n,cmp);60 61 for(i=1,num=0;i
           
          
         
        
       
      

 

哈希做法：

1 #include
      
        2 #include
       
         3 #include
        
          4 #include
         
           5 #include
          
            6 using namespace std; 7  8 const int INF = 1000; 9 typedef struct10 {11     int x,y;12 }node;13 struct hash14 {15     int x;16     int y;17     struct hash *next;18 };19 struct hash hash_table[1003];20 node a[INF+3];21 22 bool cmp(node n1,node n2)23 {24     if( n1.x!=n2.x )25         return n1.x
           
            x=x;34 new_hash->y=y;//build 35 36 new_hash->next=hash_table[k].next;37 hash_table[k].next=new_hash;38 }39 bool found(int x,int y)40 {41 unsigned k=(x*x+y*y)%INF;42 struct hash *new_hash;43 new_hash=hash_table[k].next;44 while(new_hash!=NULL)45 {46 if(new_hash->x==x && new_hash->y==y)break;47 else new_hash=new_hash->next;48 }49 if(new_hash==NULL)return false;50 else return true;51 }52 53 int main()54 {55 int n,i,j,num;56 node a1,a2,a3,a4;57 while(scanf("%d",&n)>0)58 {59 if(n==0)break;60 memset(hash_table,0,sizeof(hash_table));61 for(i=1;i<=n;i++)62 {63 scanf("%d%d",&a[i].x,&a[i].y);64 Insert(a[i].x,a[i].y);65 }66 sort(a+1,a+1+n,cmp);67 68 for(i=1,num=0;i