1058 - Parallelogram Counting PDF (English) Statistics Forum 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
平行四边形的判定定理：
两组对边分别平行的四边形是平行四边形（定义判定法）； 一组对边平行且相等的四边形是平行四边形； 两组对边分别相等的四边形是平行四边形； 两组对角分别相等的四边形是平行四边形（两组对边平行判定）； 对角线互相平分的四边形是平行四边形。 这道题用定理5判断。
记录每条边的中点坐标，如果两个边的中点坐标相同，证明这两条边为一个平行四边形的两条对角线。
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <queue> #include <stack> using namespace std; struct node { int x;int y; }; node di[1010]; node pare[1000010]; int n; bool cmp(node a,node b){ if(a.x==b.x) return a.y<b.y; return a.x<b.x; } int main(){ int T,i,j,t=1; scanf("%d",&T); while(T--){ scanf("%d",&n); for( i=1; i<=n;++i ) scanf("%d %d",&di[i].x,&di[i].y); int k=0; for( i=1;i<=n;i++){ for( j=i+1;j<=n;j++){ pare[++k].x=di[i].x+di[j].x; pare[k].y=di[i].y+di[j].y; } } sort(pare+1,pare+k+1,cmp); int temp=1; int ans=0; for( i=2;i<=k;i++){ if(pare[i].x==pare[i-1].x&&pare[i].y==pare[i-1].y) temp++; else{ ans+=temp*(temp-1)/2; temp=1; } } printf("Case %d: %d\n",t++,ans); } return 0; }