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.
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
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2 17 */ 18 #include<iostream> 19 #include<stdio.h> 20 #include<cstring> 21 #include<cstdlib> 22 #include<algorithm> 23 using namespace std; 24 25 typedef struct 26 { 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<n2.x; 34 else return n1.y<n2.y; 35 } 36 bool query(int l,int r,node cur) 37 { 38 int mid; 39 while(l<=r) 40 { 41 mid=(l+r)/2; 42 if( a[mid].x<cur.x || (a[mid].x==cur.x&&a[mid].y<cur.y)) 43 l=mid+1; 44 else if( a[mid].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<n;i++) 62 { 63 a1=a[i]; 64 for(j=i+1;j<=n;j++) 65 { 66 a2=a[j]; 67 a3.x=a1.y-a2.y+a1.x; 68 a3.y=a2.x-a1.x+a1.y; 69 if( !query(1,n,a3)) continue; 70 a4.x=a1.y-a2.y+a2.x; 71 a4.y=a2.x-a1.x+a2.y; 72 if( query(1,n,a4)) num++; 73 74 } 75 } 76 printf("%d ",num/2); 77 } 78 return 0; 79 }
哈希做法：
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 using namespace std; 7 8 const int INF = 1000; 9 typedef struct 10 { 11 int x,y; 12 }node; 13 struct hash 14 { 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<n2.x; 26 else return n1.y<n2.y; 27 } 28 void Insert(int x,int y) 29 { 30 unsigned k=(x*x+y*y)%INF; 31 struct hash *new_hash; 32 new_hash=(struct hash *)malloc(sizeof(struct hash)); 33 new_hash->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<n;i++) 69 { 70 a1=a[i]; 71 for(j=i+1;j<=n;j++) 72 { 73 a2=a[j]; 74 a3.x=a1.y-a2.y+a1.x; 75 a3.y=a2.x-a1.x+a1.y; 76 if(!found(a3.x,a3.y))continue; 77 a4.x=a1.y-a2.y+a2.x; 78 a4.y=a2.x-a1.x+a2.y; 79 if(found(a4.x,a4.y)==true) 80 num++; 81 } 82 } 83 printf("%d ",num/2); 84 } 85 return 0; 86 }