bzoj 1185 [HNOI2007]最小矩形覆盖 凸包+旋转卡壳

题目大意

用最小矩形覆盖平面上所有的点

分析

有一结论：最小矩形中有一条边在凸包的边上，不然可以旋转一个角度让面积变小
简略证明
我们逆时针枚举一条边
用旋转卡壳维护此时最左，最右，最上的点

注意

注意凸包后点数不再是n

吐槽

凸包后点数是n，bzoj上就过了？？？

solution

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
using namespace std;
typedef double db;
const db eps=1e-9;
const int M=50007;

int n;

struct pt{
	db x,y;
	pt(db _x=0.0,db _y=0.0){x=_x; y=_y;}
}p[M],s[M]; int tot;

bool eq(db x,db y){return fabs(y-x)<=eps;}
bool le(db x,db y){return eq(x,y)||x<y;}

pt operator -(pt x,pt y){return pt(x.x-y.x,x.y-y.y);}
pt operator +(pt x,pt y){return pt(x.x+y.x,x.y+y.y);}
bool operator <(pt x,pt y){return (x.y!=y.y)?(x.y<y.y):(x.x<y.x);}
bool operator ==(pt x,pt y){return eq(x.x,y.x)&&eq(x.y,y.y);};
pt operator *(pt x,db d){return pt(x.x*d,x.y*d);}
pt operator /(pt x,db d){return pt(x.x/d,x.y/d);}

db dot(pt x,pt y){
	return x.x*y.x+x.y*y.y;
}

db cross(pt x,pt y){
	return x.x*y.y-x.y*y.x;
}

db length(pt x){
	return sqrt(dot(x,x));
}

db area(pt x,pt y,pt z){
	return cross(y-x,z-x);
}

db shadow(pt x,pt y,pt to){
	return dot(y-x,to-x)/length(to-x);
}

pt lf_90(pt x){
	return pt(-x.y,x.x);
}

bool cmp(pt x,pt y){
	db tp=area(p[1],x,y);
	if(eq(tp,0)) return length(x-p[1])<length(y-p[1]);
	return tp>0;
}

void convex(){
	int i,ii=1;
 	for(i=2;i<=n;i++) if(p[i]<p[ii]) ii=i;
 	swap(p[1],p[ii]);
 	sort(p+2,p+n+1,cmp);
 	
 	s[tot=1]=p[1];
 	for(i=2;i<=n;i++){
 		while(tot>1&&le(area(s[tot-1],s[tot],p[i]),0)) tot--;
 		s[++tot]=p[i];
	 }
}

int main(){
	int i,p1,p2,p3;
	db tp1,tp2,tp3,tp4,ans;
	pt a[5],tp;
	
	scanf("%d",&n);
	
	for(i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
	
	convex();
	
	s[0]=s[tot];//要算每一条边，加上tot-0的
	
	ans=1e32;
	p1=1,p2=1,p3=1;
	
	for(i=0;i<tot;i++){
		if(s[i]==s[i+1]) continue;
		
		while(le(area(s[i],s[i+1],s[p3]),area(s[i],s[i+1],s[p3%tot+1]))) p3=p3%tot+1;
		if(i==0) p1=p3;//第一次找卡壳特例
		while(le(shadow(s[i],s[p1%tot+1],s[i+1]),shadow(s[i],s[p1],s[i+1]))) p1=p1%tot+1;
		while(le(shadow(s[i+1],s[p2%tot+1],s[i]),shadow(s[i+1],s[p2],s[i]))) p2=p2%tot+1;
		
		tp1=length(s[i+1]-s[i]);
		tp2=area(s[i],s[i+1],s[p3])/tp1;
		tp3=fabs(shadow(s[i],s[p1],s[i+1]));
		tp4=fabs(shadow(s[i+1],s[p2],s[i]));
		
		if(le((tp1+tp3+tp4)*tp2,ans)){
			ans=(tp1+tp3+tp4)*tp2;
			tp=s[i+1]-s[i];
			a[1]=s[i]-tp*(tp3/tp1);
			a[2]=s[i+1]+tp*(tp4/tp1);
			tp=lf_90(tp);
			a[3]=a[2]+tp*(tp2/tp1);
			a[4]=a[1]+tp*(tp2/tp1);
		}
	}
	
	printf("%.5lf\n",ans+eps);
	
	int ii=1;
	for(i=2;i<=4;i++) if(a[i]<a[ii]) ii=i;
	printf("%.5lf %.5lf\n",a[ii].x+eps,a[ii].y+eps);
	for(i=ii%4+1;i!=ii;i=i%4+1) printf("%.5lf %.5lf\n",a[i].x+eps,a[i].y+eps);
	
	return 0;
}