[BZOJ2338][HNOI2011]数矩形

Time Limit: 20 Sec
Memory Limit: 128 MB

Description

Sample Input

8
-2 3
-2 -1
0 3
0 -1
1 -1
2 1
-3 1
-2 1

Sample Output

10

题解：找出任意两点间的连线作为对角线，按照长度为第一关键字，中点坐标为第二关键字排序，然后在所有长度相同，中点坐标一样的对角线集合中寻找答案，因为每一个集合最多被枚举一次，那么复杂度就是O(n²)

#include
#include
#include
#include
#include
#include
#include
#include
#define LiangJiaJun main
#define pa pair
#define INF 1999122700
#define ll long long
using namespace std;
struct VEC{
    double x,y,vx,vy;
    double dis(){return vx*vx + vy*vy;}
    double cex(){return x+vx/2.0;}
    double cey(){return y+vy/2.0;}
}a[1500*1500+4];

double operator*(VEC A,VEC B){return A.vx * B.vy - A.vy * B.vx;}
double cross(VEC A,VEC B){return A.vx * B.vx + A.vy*B.vy;}
int cnt,n,x[1504],y[1504];
inline bool rosanG(VEC A,VEC B){
    return (A.cex() == B.cex())?A.cey()inline bool dex(VEC A,VEC B){
    return (A.dis() == B.dis())?rosanG(A,B):A.dis()int LiangJiaJun (){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++){
            cnt++;
            a[cnt].x=x[i];a[cnt].y=y[i];
            a[cnt].vx=x[j]-x[i];a[cnt].vy=y[j]-y[i];
        }
    sort(a+1,a+cnt+1,dex);
    int l=1,r=0;
    double ans=0;
    while(l<=cnt){
        while(a[r+1].dis() == a[l].dis()&&a[r+1].cex()==a[l].cex()&&a[r+1].cey()==a[l].cey())++r;
        for(int i=l;i<=r;i++)
            for(int j=i+1;j<=r;j++)
                ans=max(ans,fabs(a[i]*a[j]/2.0));
        l=r+1;
    }
    printf("%.0lf\n",ans);
    return 0;
}