BZOJ1185[HNOI2007] 最小矩形覆盖
题面
给定一些点的坐标,要求求能够覆盖所有点的最小面积的矩形,输出所求矩形的面积和四个顶点的坐标
分析
首先可以先求凸包，因为覆盖了凸包上的顶点，凸包内的顶点也一定能被覆盖
结论：这个矩形的一条边一定与凸包的一条边重合。
然后对于凸包的每一条边\(\vec{s_is_{i+1}}\),我们通过旋转卡壳找到最左侧的点l,最右侧的点r,最高点p,过p做\(\vec{s_is_{i+1}}\)的平行线，过l,r做\(\vec{s_is_{i+1}}\)的垂线，就得到了我们要求的矩形
求最大高度的点用叉积最大，最右侧点用点积最大，最左侧点用点积最小（点积为负）
求坐标，可以先根据前面的点积和叉积求出投影长度，再求出边对应的单位向量，乘上长度再相加就得到了点的坐标。如l的坐标就是$s_i+ \frac{\vec{s_is_{i+1}} \cdot \vec{s_il}{}}{|\vec{s_is_{i+1}}|} \frac{ \vec{s_is_{i+1}} }{|\vec{s_is_{i+1}}|} $,另两个端点只需要向量旋转90度即可
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 100000
#define eps 1e-10
#define INF 1e20
using namespace std;
int n;
struct Vector{
double x;
double y;
Vector(){
}
Vector(double _x,double _y){
x=_x;
y=_y;
}
friend Vector operator + (Vector p,Vector q){
return Vector(p.x+q.x,p.y+q.y);
}
friend Vector operator - (Vector p,Vector q){
return Vector(p.x-q.x,p.y-q.y);
}
friend Vector operator * (Vector a,double d){
return Vector(a.x*d,a.y*d);
}
friend Vector operator / (Vector a,double d){
return Vector(a.x/d,a.y/d);
}
};
typedef Vector point;
inline double dot(Vector p,Vector q){
return p.x*q.x+p.y*q.y;
}
inline double dist(point p,point q){
return sqrt(dot(p-q,p-q));
}
inline double cross(Vector p,Vector q){
return p.x*q.y-p.y*q.x;
}
inline double length(Vector x){
return sqrt(dot(x,x));
}
point a[maxn+5];
point O;
int cmp(point P,point Q){
double ang=cross(P-O,Q-O);
if(fabs(ang)<eps) return dist(O,P)<dist(O,Q);
else return ang>eps;
}
int top=0;
point s[maxn+5];
void Graham(){
for(int i=1;i<=n;i++){
if(a[i].x<a[1].x||(a[i].x==a[1].x&&a[i].y<=a[1].y)) swap(a[i],a[1]);
}
O=a[1];
sort(a+2,a+1+n,cmp);
for(int i=1;i<=n;i++){
while(top>1&&cross(s[top]-s[top-1],a[i]-s[top-1])<=eps) top--;//在逆时针方向，叉积<0
s[++top]=a[i];
}
//#ifdef DEBUG
// for(int i=1;i<=top;i++){
// printf("(%.5f,%.5f)\n",s[i].x,s[i].y);
// }
//#endif
}
inline int nex(int x){
return x%top+1;
}
inline Vector rotate90(Vector p){//把向量逆时针旋转90度
//(xcos(a)-ysin(a),xsin(a)+ycos(a)) a=pi/2
return Vector(-p.y,p.x);
}
point res[10];
double ans=INF;
void Spin(){
s[top+1]=s[1];
int l,r,p;
l=r=p=2;
for(int i=1;i<=top;i++){
double D=length(s[i+1]-s[i]);
while(cross(s[i+1]-s[i],s[p+1]-s[i])-cross(s[i+1]-s[i],s[p]-s[i])>-eps) p=nex(p);
while(dot(s[r+1]-s[i],s[i+1]-s[i])-dot(s[r]-s[i],s[i+1]-s[i])>-eps) r=nex(r);
if(i==1) l=r;
while(dot(s[l+1]-s[i],s[i+1]-s[i])-dot(s[l]-s[i],s[i+1]-s[i])<eps) l=nex(l);
//在左侧的时候点积为负数，最小
double lenl=dot(s[l]-s[i],s[i+1]-s[i])/D;
double lenr=dot(s[r]-s[i],s[i+1]-s[i])/D;
double height=cross(s[i+1]-s[i],s[p]-s[i])/D;
double area=(fabs(lenr)+fabs(lenl))*fabs(height);//lenl,lenr是有方向的
if(area<ans){
// printf("(%.3f %.3f) ",(double)s[i].x,(double)s[i].y);
// printf("(%.3f %.3f)\n",(double)s[i+1].x,(double)s[i+1].y);
// printf("S=%.4f l=%d r=%d p=%d\n",area,l,r,p);
ans=area;
res[1]=s[i]+(s[i+1]-s[i])/D*lenl;
res[2]=s[i]+(s[i+1]-s[i])/D*lenr;
res[3]=res[1]+rotate90((s[i+1]-s[i])/D)*height;
res[4]=res[2]+rotate90((s[i+1]-s[i])/D)*height;
}
}
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%lf %lf",&a[i].x,&a[i].y);
}
Graham();
Spin();
for(int i=1;i<=4;i++){
if(res[i].y<res[1].y||(res[i].y==res[1].y&&res[i].x<res[1].x)) swap(res[1],res[i]);
}
O=res[1];
sort(res+2,res+1+4,cmp);
for(int i=1;i<=4;i++){
if(fabs(res[i].x)<eps) res[i].x=0;
if(fabs(res[i].y)<eps) res[i].y=0;
}
printf("%.5f\n",ans);
for(int i=1;i<=4;i++){
printf("%.5f %.5f\n",res[i].x,res[i].y);
}
}