Lightoj 1058

题意：给n个点，问能组成的平行四边形的个数。
思路：平行四边形对角线交点唯一，算出有多少个相同的交点cnt，
sum+= （cnt*(cnt-1)）/2;

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <map>
#include <set>
#define ll long long
#define llu unsigned long long
using namespace std;
const double eps = 1e-8;
const double PI  = 3.1415926;
struct Point {
    double x,y;
    Point (double x = 0,double y = 0) : x(x),y(y) {}
};
Point p[1050],s[1000500];
struct Line{
    Point a,b;
};
Line line[100];
int indl,indp;
bool dcmp(double x) {
    if(fabs(x)-0.0 <= eps) return true;
    return false;
}

int n;
double l;
bool cmp(Point a,Point b) {
    if(!dcmp(a.x-b.x)) return a.x<b.x;
    return a.y<b.y;
}
Point A;
int main(){
    int t; scanf("%d",&t);
    for(int cas = 1; cas<=t; cas++){
        int indx = 1;
        scanf("%d",&n);
        for(int i=1;i<=n; i++) {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            for(int j=1; j<i; j++) {
                s[indx].x = (p[i].x+p[j].x)/2;
                s[indx].y = (p[i].y+p[j].y)/2;
                indx++;
            }
        }
        ll sum = 0;
        ll ans = 1;
        sort(s+1,s+indx,cmp);
        A.x = s[1].x; A.y = s[1].y;
        for(int i=2;i<indx;i++) {
            if(dcmp(s[i].x-A.x) && dcmp(s[i].y-A.y)) {
                ans++;
            }else {
                if(ans >= 2) {
                    sum += (ans*(ans-1)/2);
                }
                ans = 1;
                A.x = s[i].x; A.y = s[i].y;
            }
        }
        printf("Case %d: %lld\n",cas,sum);
    }
    return 0;
}