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 nextn lines, contains2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than1000000000.
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
利用平行四边形的性质,两对角线的中点相等来做,可以把每两个点的中点都求出来,然后排序找相等的点,相加即可;
#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define INF 0xfffffff
using namespace std;
struct point
{
int x,y;
};
point p[1100],a[500100];
int cmp(point a,point b)
{
if(a.x!=b.x)
{
return a.x<b.x;
}
else
{
return a.y<b.y;
}
}
int main()
{
int t,n,i,j,ans,count;
int T=0;
scanf("%d",&t);
while(t--)
{
memset(p,0,sizeof(p));
memset(a,0,sizeof(a));
scanf("%d",&n);
for(i=0; i<n; i++)
scanf("%d%d",&p[i].x,&p[i].y);
int k=0;
for(i=0; i<n; i++)
for(j=i+1; j<n; j++)
{
a[k].x=1.0*(p[i].x+p[j].x);
a[k].y=1.0*(p[i].y+p[j].y);
k++;
}
sort(a,a+k,cmp);//这里注意排序要写快排,用冒泡会超时,快排应该用sort,用c语言的qsort无法进行正确的排序
ans=1;
count=0;
for(i=0; i<k-1; i++)
{
if(a[i].x==a[i+1].x&&a[i].y==a[i+1].y)
ans++;
else
{
count+=(ans-1)*ans/2;
ans=1;
}
}
count+=(ans-1)*ans/2;
printf("Case %d: %d\n",++T,count);
}
return 0;
}