Squares<哈希>

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



就是看看给出的点能组成几个正方形；
不要四个四个的组合，想也会超时，任意两个组合，然后计算出另外两个点；
再去查询是否存在；结果要除以4的；

 1 #include<iostream>

 2 #include<cstring>

 3 #include<cstdio>

 4 using namespace std;

 5 bool x[5000][5000];//我才不告诉你其实数据是-2000--2000

 6 int si[1010],sj[1010];

 7 int f(int a,int b)

 8 {

 9     if(x[a+2500][b+2500])

10         return 1;

11     return 0;

12 }

13 int main()

14 {

15     int a,b,i,j,n,sum;

16     int x1,y1,x2,y2,x3,y3,x4,y4;

17     while(scanf("%d",&n)&&n)

18     {

19         sum=0;

20         memset(x,0,sizeof(x));

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

22         {

23             scanf("%d %d",&a,&b);

24             si[i]=a;

25             sj[i]=b;

26             x[a+2500][b+2500]=1;

27         }

28         for(i=1; i<n; i++)

29         {

30             x1=si[i];

31             y1=sj[i];

32             for(j=0; j<i; j++)

33             {

34                 x2=si[j];

35                 y2=sj[j];

36                 x3=x1+(y1-y2);//计算剩下两个点

37                 y3= y1-(x1-x2);

38                 x4=x2+(y1-y2);

39                 y4= y2-(x1-x2);

40                 if(f(x3,y3)&&f(x4,y4))

41                     sum++;

42                 x3=x1-(y1-y2);

43                 y3= y1+(x1-x2);

44                 x4=x2-(y1-y2);

45                 y4= y2+(x1-x2);

46                 if(f(x3,y3)&&f(x4,y4))

47                     sum++;

48             }

49         }

50         printf("%d\n",sum/4);//记得除以4

51     }

52     return 0;

53 }

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