D. Number of Parallelograms
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time limit per test
4 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputYou are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
input
40 11 01 12 0
output
1
主要思路就是把点转换为向量,平行四边形的两条对边的向量是相等的。
而又有2条对边,并且向量是有方向的,一正一负,这样相同的就又多了2对,所以结果就除以4.
#include <cstdio>#include <map>#include <iostream>using namespace std;int x[2222];int y[2222];map<pair<int, int>, int> cnt;int main() { int n; cin >> n; for (int i = 1; i <= n; i++) { cin >> x[i] >> y[i]; } int ans = 0; for (int i = 1; i <= n; i++) { for (int j =1; j <= n; j++) { if(i==j)continue; int dx = x[i] - x[j]; int dy = y[i] - y[j]; ans += cnt[make_pair(dx, dy)]++; } } cout << ans / 4; return 0;}
可能有同学会想,我把for循环的第2重改为for(int j=i+1;j<=n;j++)这样就去掉重复的正负向量了,然后ans/2即可。
然而这样是不可以的,我之前代码就是这么写的,结果就跪了。
如果把输入的点的顺序按照1,2,3,4,5....的顺序排列。
而平行四边形,可能向量12,43是对边,也可能向量12,34是对边。
如果是那样写的话,后一个的点的数字一定是大于i的,也就是12,43这种情况就不能满足了。
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