poj 2002 Squares

题目链接：http://poj.org/problem?id=2002

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.

题意：二维平面给出一些点，计算以这些点为顶点的正方形的个数。

解法：哈希。因为n为1000，坐标大小高达20000，普通查找肯定不行的，我们得用哈希压缩坐标。

　　   然后枚举正方形的一条边上的两个顶点，可以计算出剩余的两个顶点坐标，然后哈希表查找是否有这样的正方形，统计个数，然后除以4即可（因为枚举了每个正方形的每条边）。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<cstdlib>

 5 #include<cmath>

 6 #include<algorithm>

 7 #define inf 0x7fffffff

 8 using namespace std;

 9 typedef long long ll;

10 const int maxn=1000+10;

11 const int mod=100007;

12 struct node

13 {

14     int x,y;

15     int next;

16 }edge[maxn];

17 int head[mod],edgenum;

18 int xx[maxn],yy[maxn];

19 void insert_hash(int x,int y)

20 {

21     int k=(x*x+y*y)%mod;

22     edge[edgenum].x=x ;

23     edge[edgenum].y=y ;

24     edge[edgenum].next=head[k] ;

25     head[k]=edgenum++ ;

26 }

27 int search_hash(int x,int y)

28 {

29     int k=(x*x+y*y)%mod;

30     for (int i=head[k] ;i!=-1 ;i=edge[i].next)

31     {

32         if (x==edge[i].x && y==edge[i].y) return true;

33     }

34     return false;

35 }

36 int main()

37 {

38     int n;

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

40     {

41         memset(head,-1,sizeof(head));

42         edgenum=0;

43         for (int i=0 ;i<n ;i++)

44         {

45             scanf("%d%d",&xx[i],&yy[i]);

46             insert_hash(xx[i],yy[i]);

47         }

48         int count=0;

49         for (int i=0 ;i<n ;i++)

50         {

51             for (int j=i+1 ;j<n ;j++)

52             {

53                 int x=xx[i]+(yy[j]-yy[i]);

54                 int y=yy[i]-(xx[j]-xx[i]);

55                 int x2=xx[j]-(yy[i]-yy[j]);

56                 int y2=yy[j]+(xx[i]-xx[j]);

57                 if (search_hash(x,y) && search_hash(x2,y2)) count ++ ;

58             }

59         }

60         for (int i=0 ;i<n ;i++)

61         {

62             for (int j=i+1 ;j<n ;j++)

63             {

64                 int x=xx[i]-(yy[j]-yy[i]);

65                 int y=yy[i]+(xx[j]-xx[i]);

66                 int x2=xx[j]+(yy[i]-yy[j]);

67                 int y2=yy[j]-(xx[i]-xx[j]);

68                 if (search_hash(x,y) && search_hash(x2,y2)) count ++ ;

69             }

70         }

71         printf("%d\n",count/4);

72     }

73     return 0;

74 }

View Code

 

 

后续：感谢大牛提出宝贵的意见。。。