POJ 2002 Squares

题目：

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

要解这道题需要的数学知识:
已知：p1=（x1,y1）,p2=(x2,y2)找在坐标轴上所对应的正方形的的另外两个坐标p3=（x3,y3）,p4=（x4,y4）。
则有公式：
         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)

先枚举出两个点p1,p2。在通过数学公式求解出p3,p4点，再用二分法看所给点中是否有p3,p4这两个点。
看了别人的解法，还有用hash的，但目前还不会~= =

代码：

 1 #include<iostream>

 2 #include<algorithm>

 3 #include<cstdio>

 4 using namespace std;

 5 

 6 int n;

 7 

 8 class Point

 9 {

10 public:

11     int x,y;

12 }point[1100];

13 

14 bool cmp(Point a,Point b)

15 {

16     if(a.x==b.x)

17         return a.y<b.y;

18     return a.x<b.x;

19 }

20 

21 int judge(int a,int b)

22 {

23     int left=0,right=n-1,mid;

24     while(left<=right)

25     {

26         mid=(left+right)/2;

27         if(point[mid].x==a && point[mid].y==b)

28                return 1;

29         else

30             if(point[mid].x<a || (point[mid].x==a && point[mid].y<b))

31                 left=mid+1;

32             else

33                 right=mid-1;

34     }

35     return 0;

36 }

37 

38 int main()

39 {

40     int i,j;

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

42     {

43         int sum=0,x,y,i,j;

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

45             scanf("%d%d",&point[i].x,&point[i].y);

46         sort(point,point+n,cmp);

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

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

49         {

50             x=point[i].y-point[j].y+point[i].x;        //这四个求坐标的公式很重要。。。。。

51             y=point[j].x-point[i].x+point[i].y;

52             if(!judge(x,y)) continue;

53             x=point[i].y-point[j].y+point[j].x;

54             y=point[j].x-point[i].x+point[j].y;

55             if(judge(x,y)) sum++;

56         }

57         printf("%d\n",sum/2);

58     }

59     return 0;

60 }