[BZOJ1185][HNOI2007]最小矩形覆盖-[凸包+旋转卡壳]

Description

传送门

Solution

感性理解一下，最小矩形一定是由一条边和凸包上的边重合的。

然后它就是模板题了。。然而真的好难调，小于大于动不动就打错。

Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const double eps=1e-;
int n;
struct node{double x,y;
    friend node operator +(node a,node b){return node{a.x+b.x,a.y+b.y};}
    friend node operator -(node a,node b){return node{a.x-b.x,a.y-b.y};}
    friend double operator *(node a,node b){return a.x*b.y-a.y*b.x;}
    friend double operator /(node a,node b){return a.x*b.x+a.y*b.y;}
    friend node operator *(node a,double b) {return node{a.x*b,a.y*b};}
    friend bool operator <(node a,node b) { return fabs(a.y-b.y)<eps?a.x<b.x:a.y<b.y;}
}p[],st[],low;
double dis(node a){return a.x*a.x+a.y*a.y;}
bool cmp(node a,node b){
    double t=(a-p[])*(b-p[]);
    if (fabs(t)<eps) return dis(a-p[])<dis(b-p[]);
    return t>;
}
int top=,now;
int main()
{
    scanf("%d",&n);
    low.x=low.y=;
    scanf("%lf%lf",&p[].x,&p[].y);
    for (int i=;i<=n;i++)
    {
        scanf("%lf%lf",&p[i].x,&p[i].y);
        if (p[i]<p[]) swap(p[i],p[]);
    }
    sort(p+,p+n+,cmp);
    st[++top]=p[];
    for (int i=;i<=n;i++)
    {
        while (top>&&(st[top]-st[top-])*(p[i]-st[top])<eps) top--;
        st[++top]=p[i];
    }
    st[]=st[top];
    node pr[];
    double L,R,D,H,ans=1e10;int l=,r=,now=;
    for (int i=;i<top;i++)
    {
        D=sqrt(dis(st[i]-st[i+]));
        while ((st[i+]-st[i])*(st[now+]-st[i])>(st[i+]-st[i])*(st[now]-st[i])-eps) now=(now+)%top;
        while ((st[i+]-st[i])/(st[r+]-st[i])>(st[i+]-st[i])/(st[r]-st[i])-eps) r=(r+)%top;
        if (!i) l=r;
        while ((st[i+]-st[i])/(st[l+]-st[i])<(st[i+]-st[i])/(st[l]-st[i])+eps) l=(l+)%top;
        L=(st[i+]-st[i])/(st[l]-st[i])/D;R=(st[i+]-st[i])/(st[r]-st[i])/D;
        H=abs((st[i+]-st[i])*(st[now]-st[i])/D);
        if ((R-L)*H<ans)
        {
            ans=(R-L)*H;
            pr[]=st[i]+(st[i+]-st[i])*(R/D);
            pr[]=pr[]+(st[r]-pr[])*(H/sqrt(dis(pr[]-st[r])));
            pr[]=pr[]-(pr[]-st[i])*((R-L)/sqrt(dis(st[i]-pr[])));
            pr[]=pr[]-(pr[]-pr[]);
        }
    }
    now=;
    for (int i=;i<=;i++) if (pr[i]<pr[now]) now=i;
    printf("%.5f\n",ans);
    for (int i=;i<=;i++)
    {
        if (pr[(i+now)%].x>-*1e-) pr[(i+now)%].x=fabs(pr[(i+now)%].x);
        if (pr[(i+now)%].y>-*1e-) pr[(i+now)%].y=fabs(pr[(i+now)%].y);
        printf("%.5f %.5f\n",pr[(i+now)%].x,pr[(i+now)%].y);
    }
}