POJ
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C - Parallelogram Counting
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
#include<stdio.h>
#include<string.h>
#include<cstdio>
#include<map>
#include<iostream>
#include<algorithm>
#define M 10000000
using namespace std;
struct hanshu
{
int x,y;
}a[M];
struct zhi
{
int x,y;
}b[M];
int cmp(struct zhi q,struct zhi w)
{
if(q.x!=w.x)
{
return q.x<w.x;
}
else
{
return q.y<w.y;
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int t,i,j;
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
}
int k=0;
for(i=0;i<t;i++)
{
for(j=i+1;j<t;j++)
{
int x=(a[i].x+a[j].x);
int y=(a[i].y+a[j].y);
b[k].x=x,b[k].y=y;
k++;
}
}
sort(b,b+k,cmp);
int sum=0,c=1;
int x1=b[0].x,y1=b[0].y;
for(i=1;i<k;i++)
{
if(b[i].x==x1&&b[i].y==y1)
{
c++;
}
else
{
sum=sum+(c)*(c-1)/2;
x1=b[i].x,y1=b[i].y;
c=1;
}
}
if(c!=1)
{
sum=sum+(c)*(c-1)/2;
}
printf("%d\n",sum);
}
return 0;
}
判断能组成多少个平行四边形。
这题利用平行四边形的性质中点到对角线的距离相等来求然后排序。
