[BZOJ2338][HNOI2011]数矩形

Time Limit: 20 Sec
Memory Limit: 128 MB

Description

Sample Input

8-2 3-2 -10 30 -11 -12 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<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#include<queue>#include<vector>#define LiangJiaJun main#define pa pair<int,int>#define INF 1999122700#define ll long longusing 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()<B.cey():A.cex()<B.cex();}inline bool dex(VEC A,VEC B){    return (A.dis() == B.dis())?rosanG(A,B):A.dis()<B.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;}