Squares(哈希)

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 14328		Accepted: 5393

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
题意：给出n个点的坐标，计算这些点可以构成多少个正方形，不同顺序的相同四个点被视为同一个正方形。

思路：这里n最大是1000，显然一个点一个点的枚举不行。参考了别人的结题报告，据说有这样的定理：
　　　已知（x1,y1） 和 （x2,y2）
     则 
     x3 = x1 + (y1-y2); y3 = y1 - (x1-x2);
     x4 = x2 +（y1-y2;  y4 = y2 - (x1-x2);
     或
     x3 = x1 - (y1-y2); y3 = y1 + (x1-x2);
     x4 = x2 - (y1-y2); y4 = y2 + (x1-x2);
     因此可以先枚举两个点，根据这两个点（点1 ，点2）的坐标可以得到另外两个点（点3， 点4），若在哈希表中能找得到这两个（点3， 点4），说明能构成一个正方形，
     注意每个点被枚举了四次，最后要除以4；
     再者就是找哈希函数，这里用的平方取余法，每输入一个点，就将这个点插入哈希表中；
     这个题也受了 poj 3274的启发；

 1 #include<stdio.h>

 2 #include<string.h>

 3 

 4 const int prime = 99991;

 5 struct node

 6 {

 7     int x,y;

 8 }pos[1010];

 9 

10 struct HashTable

11 {

12     int x;

13     int y;

14     struct HashTable* next;

15 }*Hash[prime];//Hash[]是指针数组，存放地址；

16 int n;

17 

18 //插入哈希表

19 void hash_insert(int x, int y)

20 {

21     int key = (x*x + y*y)%prime;//平方求余法；

22     if(!Hash[key])

23     {

24         Hash[key] = new struct HashTable;

25         Hash[key]->x = x;

26         Hash[key]->y = y;

27         Hash[key]->next = NULL;

28     }

29     else

30     {

31         struct HashTable *tmp = Hash[key];

32         while(tmp->next)

33             tmp = tmp->next;//开放寻址，直到next为空

34         //插入新结点

35         tmp->next = new struct HashTable;

36         tmp->next->x = x;

37         tmp->next->y = y;

38         tmp->next->next = NULL;

39     }

40 }

41 bool find(int x, int y)

42 {

43     int key = (x*x+y*y)%prime;

44     if(!Hash[key])

45         return false;//key 对应的地址不存在，

46     else

47     {

48         struct HashTable *tmp = Hash[key];

49         while(tmp)

50         {

51             if(tmp->x == x && tmp->y == y)

52                 return true;

53             tmp = tmp->next;

54         }

55         return false;

56     }

57 }

58 int main()

59 {

60     while(scanf("%d",&n)!= EOF)

61     {

62         int i,j;

63         if(n == 0) break;

64         memset(Hash,0,sizeof(Hash));

65         for(i = 0; i < n; i++)

66         {

67             scanf("%d %d",&pos[i].x,&pos[i].y);

68             hash_insert(pos[i].x, pos[i].y);

69         }

70         int ans = 0;

71         for(i = 0; i < n-1; i++)

72         {

73             for(j = i+1; j < n; j++)

74             {

75                 int x1 = pos[i].x, y1 = pos[i].y;

76                 int x2 = pos[j].x, y2 = pos[j].y;

77                 int add_x = x1-x2,add_y = y1-y2;

78                 int x3 = x1+add_y;

79                 int y3 = y1-add_x;

80                 int x4 = x2+add_y;

81                 int y4 = y2-add_x;

82                 if(find(x3,y3) && find(x4,y4))

83                     ans++;

84 

85                  x3 = x1-add_y;

86                  y3 = y1+add_x;

87                  x4 = x2-add_y;

88                  y4 = y2+add_x;

89                 if(find(x3,y3) && find(x4,y4))

90                     ans++;

91             }

92         }

93         printf("%d\n",ans/4);

94     }

95     return 0;

96 }

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