lightoj1058 Parallelogram Counting

思路：一个二维坐标系中给出n个点，可以两两连线，问这些所有线段中能组成多少平行四边形。n <= 1000。

这里显然不能枚举组合，那样是n^4的做法，必然是超时的。那么我们可以用平四边形的等价定义，两条相互平分的线段的四个点是平行四边形的顶点，那么我们可以先用n^2de方法求出任意两个点的连线（某四边形的duijiaoxian）的中点。这样就有n*(n-1)/2个中点。然后我们sort一下，求出相同的点的个数，然后一这个点某平行四边形对角线的中点的四边形个数为x(x-1)/2。这样整体的时间复杂度是n^2*log(n)（排序）。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define CLR(x) memset(x, 0,sizeof x)
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
template<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}
template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Point{
double x, y;
Point(){}
Point(double _x, double _y){
x = _x;
y = _y;
}
bool operator < (const Point& rhs)const{
return x < rhs.x || (x == rhs.x && y < rhs.y);
}
Point operator + (const Point& rhs)const{
return Point(x + rhs.x, y + rhs.y);
}
Point operator - (const Point& rhs)const{
return Point(x - rhs.x, y - rhs.y);
}
double operator ^ (const Point& rhs)const{
return (x * rhs.y - y * rhs.x);
}
double operator * (const Point& rhs)const{
return (x * rhs.x + y * rhs.y);
}
Point operator * (const double Num)const{
return Point(x * Num,y * Num);
}
friend ostream& operator << (ostream& output,const Point& rhs){
output << "(" << rhs.x << "," << rhs.y << ")";
return output;
}
};
typedef Point Vector;
/*向量的模*/
inline double Length(const Vector& A){//Get the length of vector A;
return sqrt(A * A);//sqrt(x*x+y*y);
}
/*向量夹角*/
inline double Angle(const Vector& A,const Point& B){
return acos(A * B/Length(A)/Length(B));
}
/*向量A旋转rad弧度*/
inline Point Rotate(const Vector& A, double rad){
return Vector(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad));
}
Point p[1010],ans[1000*500];
int main()
{	// ios::sync_with_stdio(false);
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t, icase = 0, n;
cin >> t;
while(t--){
cin >> n;
for (int i = 0;i < n;++i)
cin >> p[i].x >> p[i].y;
LL sum = 0;
int k = 0;
for (int i = 0;i < n;++i){
for (int j = i + 1;j < n;++j){
ans[k].x = p[i].x + p[j].x;// ans[k].x /= 2;
ans[k].y = p[i].y + p[j].y;// ans[k].x /= 2;
k++;
}
}
sort(ans,ans + k);
int j;
for (int i = 0;i < k;++i){
for (j = i + 1;j < k;++j){
if (ans[i].x != ans[j].x || ans[i].y != ans[j].y) break;
}
n = j - i;
sum += 1LL*(n*(n - 1)) / 2;
i = j - 1;
}
printf("Case %d: %lld\n", ++icase, sum);
}
return 0;
}