标签：std lan problem ams show ble cout ++ poi
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 file 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
Source: Tehran 2004 Iran
Nationwide
平行四边形两对角线的中点重合。求平行四边形数即求相等的中点对数。
#include <iostream> #include <algorithm> using namespace std; struct vertice{ int x,y; }; vertice p[1010]; vertice q[1000010]; int cmp(vertice a,vertice b) { if(a.x==b.x) { return a.y<b.y; } return a.x<b.x; } int main(){ int t,n,i,j; cin>>t; while(t--){ cin>>n; for(i=0;i<n;i++) { cin>>p[i].x>>p[i].y; } int c = 0; for(i=0;i<n;i++) { for(j=i+1;j<n;j++) { q[c].x=p[i].x+p[j].x;//求中点未除2 q[c].y=p[i].y+p[j].y; c++; } } sort(q,q+c,cmp); long long num = 0; long long cnt = 1; int x = q[0].x; int y = q[0].y; for(i=1;i<c;i++){ if(q[i].x==x&&q[i].y==y){ cnt++; } else{ x = q[i].x; y = q[i].y; num += (cnt-1)*cnt/2; cnt = 1; } } if(cnt!=1) { num += (cnt)*(cnt-1)/2; } cout<<num<<endl; } return 0; }
TOJ-1313 Parallelogram Counting
标签：std lan problem ams show ble cout ++ poi
原文地址：http://www.cnblogs.com/shenchuguimo/p/6337893.html
