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| 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 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 integersx and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000.
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 |
题解：
给一系列点，让找到可以组成的平行四边形的个数；
思路：由于平行四边形的交点是唯一的，那么我们只要记录每两个直线的交点，判断交点重复的个数可以求出来了；假设有这个交点有三个重复的，则ans+=3*2/2;
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; const int INF=0x3f3f3f3f; typedef long long LL; const int MAXN=1010; struct Dot{ double x,y; bool operator < (const Dot &b) const{ if(x!=b.x)return x<b.x; else return y<b.y; } }; Dot dt[MAXN]; Dot num[MAXN*MAXN]; int main(){ int T,N,kase=0; scanf("%d",&T); while(T--){ scanf("%d",&N); for(int i=0;i<N;i++)scanf("%lf%lf",&dt[i].x,&dt[i].y); int tp=0; for(int i=0;i<N;i++) for(int j=i+1;j<N;j++){ if(dt[i].x==dt[j].x&&dt[i].y==dt[j].y)continue; num[tp].y=(dt[i].y+dt[j].y)/2; num[tp].x=(dt[i].x+dt[j].x)/2; tp++; } sort(num,num+tp); LL ans=0,cur=1; for(int i=1;i<tp;i++){ if(num[i].x==num[i-1].x&&num[i].y==num[i-1].y)cur++; else{ ans+=cur*(cur-1)/2;cur=1; } } printf("Case %d: %lld\n",++kase,ans); } return 0; }
Parallelogram Counting（平行四边形个数，思维转化）
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原文地址：http://www.cnblogs.com/handsomecui/p/5259374.html
