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1058 - Parallelogram Counting    PDF (English)StatisticsForum Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for 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

Case 1: 5

Case 2: 6

差点wa到放弃比赛的题。刚开始思路是先选出所有平行的线，然后去掉三点共线的，然后除以2就得到结果，但是还是TLE了。

后来把线的向量全部放到一四象限（这一点出了小问题，查了俩小时），然后再计数的时候同时判断三点共线，这样就省时间了，最后终于查出来了。

代码如下：

#include <stdio.h> #include <cstring> #include <algorithm> using namespace std; #define CLR(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define LL long long struct node { int x,y; }dot[1011]; struct node2 { int a,b; int x,y; }edge[500500]; bool check(node2 a,node2 b) { if (a.a == b.a || a.a == b.b || a.b == b.a || a.b == b.b) return false; //三点共线 if ((dot[b.b].x - dot[a.a].x) * a.y == (dot[b.b].y - dot[a.a].y) * a.x) return false; return true; } bool cmp(node2 a,node2 b) { if (a.x == b.x) return a.y < b.y; return a.x < b.x; } int main() { int u; int n; int Case = 1; scanf ("%d",&u); while (u--) { scanf ("%d",&n); for (int i = 0 ; i < n ; i++) scanf ("%d %d",&dot[i].x,&dot[i].y); int ans = 0; int ant = 0; for (int i = 0 ; i < n ; i++) { for (int j = i + 1 ; j < n ; j++) { edge[ant].x = dot[i].x - dot[j].x; edge[ant].y = dot[i].y - dot[j].y; edge[ant].a = i; edge[ant].b = j; if (edge[ant].x <= 0 && edge[ant].y <= 0) { edge[ant].x = -edge[ant].x; edge[ant].y = -edge[ant].y; } if (edge[ant].x <= 0 && edge[ant].y >= 0) //这里y的值可能由第一个x的值改变 { edge[ant].x = -edge[ant].x; edge[ant].y = -edge[ant].y; } ant++; } } sort(edge , edge+ant , cmp); int st = 0; while (st < ant) { int i,j; for (i = st ; edge[i].x == edge[st].x && edge[i].y == edge[st].y ; i++) { for (j = i + 1 ; edge[j].x == edge[st].x && edge[j].y == edge[st].y ; j++) { if (check(edge[i],edge[j])) ans++; } } st = i; } ans >>= 1; printf ("Case %d: %d\n",Case++,ans); } return 0; }