LIght OJ 1058

H - H

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1058 uDebug

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

Input starts with an integer T (≤ 15), denoting the number of test cases.

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

For each case, print the case number and the number of parallelograms that can be formed.

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

Case 1: 5

Case 2: 6

题意： 计算平行四边形的个数，，，，，

思路：用结构体记录任意两个点之间的x坐标和与y坐标和，(相当于这两个点作为一条对角线的情况），然后找出对角线相交于同一点的个数k（结构体中的x和y相等），然后C(k,2)就OK,因为若多条对角线相交于同一个中点，则任意选择两条就可以构成一个平行四边形；

代码：

#include<stdio.h> #include<string.h>#include<algorithm> using namespace std;int x[1001];int y[1001];struct node{int x;int y;}a[1000*1000/2];int cmp(node a,node b){if(a.x!=b.x){return a.x<b.x;}elsereturn a.y<b.y;}int main(){int t,n,mm=1;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d%d",&x[i],&y[i]);}int t=0;for(int i=0;i<n;i++){for(int j=i+1;j<n;j++){a[t].x=x[i]+x[j];a[t].y=y[i]+y[j];t++;}}int num=0,ans=1,cnt=0;sort(a,a+t,cmp);for(int i=1;i<t;i++){if(a[i].x==a[num].x&&a[i].y==a[num].y){ans++;//记录经过一个中点的对角线的个数； }else{cnt+=ans*(ans-1)/2;//计算经过这个中点的平行四边形的个数； num=i;//记录经过下一个中点的一条对角线，作为后面比较的对象； ans=1;//每次计算经过不同中点的对角线个数时，都要初始化为1； }}printf("Case %d: %d\n",mm++,cnt);}return 0;}