1058 - Parallelogram Counting 计算几何
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
|
分析：学了这么多年数学，然而只知道平行四边形对角线交于一点，却没想到只要存在两点的中点与另两点的中点相同，就能构成平行四边形。
代码：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
#define N 510000
struct node
{
int x;
int y;
}arr[N], mid[N];
bool cmp(struct node a, struct node b)
{
if(a.x != b.x)
return a.x < b.x;
return a.y < b.y;
}
int main(void)
{
int T, cas, n;
scanf("%d", &T);
cas = 0;
while(T--)
{
cas++;
memset(mid, 0, sizeof(mid));
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%d%d", &arr[i].x, &arr[i].y);
}
int num = 0;
for(int i = 0; i < n-1; i++)
{
for(int j = i+1; j < n; j++)
{
mid[num].x = arr[i].x + arr[j].x;
mid[num].y = arr[i].y + arr[j].y;
num++;
}
}
sort(mid, mid+num, cmp);/// 排序是为了方便下面比较
int cnt = 1;
int ans = 0;
int flag = 0;
for(int i = 1; i < num; i++)
{
if(mid[i].x == mid[flag].x && mid[i].y == mid[flag].y)
{
cnt++;
}
else
{
ans += cnt * (cnt - 1) / 2;///计算就是组合数：从 n个数里面取出 2个数 ，就是 C(n,2)
cnt = 1;
flag = i;
}
}
if(cnt>1)
ans += (cnt - 1) * cnt / 2; /// 判断循环的最后一组数据，如果也存在相同的点，就加上
printf("Case %d: %d\n", cas, ans);
}
return 0;
}
