POJ---1791 Parallelogram Counting[数学题-平行四边形求个数]

Parallelogram Counting

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5361		Accepted: 1794

Description

There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points that can be written as {A, B, C, D} such that AB || CD, and BC || AD. No four points are in a straight line.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases. It is followed by the input data for each test case.
The first line of each test case contains an integer n (1 <= n <= 1000). Each of the next n lines, contains 2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000.

Output

Output should contain t lines.
Line i contains an integer showing the number of the parallelograms as described above for test case i.

Sample Input

2

6

0 0

2 0

4 0

1 1

3 1

5 1

7

-2 -1

8 9

5 7

1 1

4 8

2 0

9 8

Sample Output

5

6

Source

Tehran Sharif 2004 Preliminary

 

 

 

 

【题目大意】：给出n个点，求出这n个点能够组成平行四边形的个数。

【解题思路】：

1）平行四边形的对角线的中点一定相交。<=> 如果有两条不同线段的中点相交，就是一个平行四边形

2）利用点坐标求出中点的集合，离散化后求出同个中点的出现的个数k。
3）对于每一个k ，利用组合公式C(k,2)的答案就是平行四边行的个数

 

 

code:

 1 #include<iostream>

 2 #include<algorithm>

 3 using namespace std;

 4 

 5 #define MAXN 1010

 6 

 7 typedef struct point

 8 {

 9     int x,y;

10 }Point;

11 Point point[MAXN];

12 Point mid[MAXN*MAXN];

13 

14 int cmp(const Point &a,const 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 main()

22 {

23     int t;

24     int i,j;

25     int sum;

26     int n;

27     int cnt;

28     scanf("%d",&t);

29     while(t--)

30     {

31         scanf("%d",&n);

32         sum=0;

33         cnt=0;

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

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

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

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

38             {

39                 mid[cnt].x=(point[i].x+point[j].x);

40                 mid[cnt].y=(point[i].y+point[j].y);

41                 cnt++;

42             }

43         sort(mid,mid+cnt,cmp);

44         int count=1;

45         for(i=0;i<cnt;i++)

46         {

47             if(mid[i].x==mid[i+1].x&&mid[i].y==mid[i+1].y)

48                 count++;

49             else

50             {

51                 sum+=(count-1)*count/2;

52                 count=1;

53             }

54         }

55         printf("%d\n",sum);

56     }

57     return 0;

58 }