思路:

首先有一个显然的性质:就是最小矩形必定有一条边与凸包的边重合.

然后我们自然就是对于每一条边考虑一下:首先找到距离这条边最远的点.然后从这个点到这条边做一条垂线段,用这条垂线段上下平移来得到两个边界.

于是我用旋转卡壳找最远的点,然后对于两边分别三分.

效率好像还不低.

#include<cstdio>
#include<cstring>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
 
typedef double f2;
 
static const f2 eps=1e-7;
inline int dcmp(const f2&x){return fabs(x)<eps?0:(x<0?-1:1);}
 
#define N 50010
struct Point{
    f2 x,y;
    Point(){}
    Point(f2 _x,f2 _y):x(_x),y(_y){}
     
    inline void read(){scanf("%lf%lf",&x,&y);}
    inline void print(){printf("%.5lf %.5lf\n",x,y);}
     
    bool operator<(const Point&B)const{return fabs(x-B.x)==0?y<B.y:x<B.x;}
     
    Point operator+(const Point&B)const{return Point(x+B.x,y+B.y);}
    Point operator-(const Point&B)const{return Point(B.x-x,B.y-y);}
    Point operator*(const f2&p)const{return Point(x*p,y*p);}
};
 
Point getvec(const Point&v){return Point(-v.y,v.x);}
Point getmid(const Point&a,const Point&b){return Point((a.x+b.x)/2,(a.y+b.y)/2);}
 
f2 getdis(const Point&a,const Point&b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
f2 cross(const Point&a,const Point&b){return a.x*b.y-a.y*b.x;}
f2 area(const Point&a,const Point&b,const Point&c){return fabs(cross(a-b,a-c))*.5;}
 
struct Line{
    Point p,v;
    Line(Point p1,Point p2,bool d){p=p1,v=d?p1-p2:p2;}
};
 
Point getlineintersection(const Line&l1,const Line&l2){
    return l1.p+l1.v*(cross(l1.p-l2.p,l2.v)/cross(l1.v,l2.v));
}
 
Point P[N],stk[N<<1];int top;
 
Point move(Point p,Point v,f2 l){
    return p+v*(l/sqrt(v.x*v.x+v.y*v.y));
}
 
f2 res=1e10;
Point re[N];
 
int main(){
    //freopen("tt.in","r",stdin);
    int n;scanf("%d",&n);
    register int i,j;
     
    for(i=1;i<=n;++i)P[i].read();
     
    sort(P+1,P+n+1);
    int ins;for(ins=n;ins!=1&&dcmp(P[ins].x-P[ins-1].x)==0;--ins);
    for(i=ins;i<=(n+ins)/2;++i)swap(P[i],P[n+ins-i]);
     
    for(i=1;i<=n;++i){
        if(!top)stk[++top]=P[i];
        else{while(top>=2&&dcmp(cross(stk[top-1]-stk[top],stk[top]-P[i]))<=0)--top;stk[++top]=P[i];}
    }
    int instk=top;
    for(i=ins-1;i>=1;--i){
        if(top==instk)stk[++top]=P[i];
        else{while(top>=instk+1&&dcmp(cross(stk[top-1]-stk[top],stk[top]-P[i]))<=0)--top;stk[++top]=P[i];}
    }--top;
     
    for(i=top+1;i<=2*top;++i)stk[i]=stk[i-top];
     
    int r=2,L,R,ll,rr;
    f2 h1,h2;
    for(i=1;i<=top;++i){
        while(area(stk[i],stk[i+1],stk[r])<=area(stk[i],stk[i+1],stk[r+1]))++r;
         
        Line l1=Line(stk[i],stk[i+1],1);
        Line l2=Line(stk[r],getvec(l1.v),0);
        Point pr=getlineintersection(l1,l2);
         
        h1=h2=0;
        for(L=i+1,R=r;;){
            if(R-L+1<=5){for(j=L;j<=R;++j)h1=max(h1,area(stk[r],pr,stk[j])*2/getdis(stk[r],pr));break;}
             
            ll=L+(R-L+1)/3,rr=R-(R-L+1)/3;
            if(area(stk[r],pr,stk[ll])<=area(stk[r],pr,stk[rr]))L=ll;else R=rr;
        }
        for(L=r,R=i+top;;){
            if(R-L+1<=5){for(j=L;j<=R;++j)h2=max(h2,area(stk[r],pr,stk[j])*2/getdis(stk[r],pr));break;}
             
            ll=L+(R-L+1)/3,rr=R-(R-L+1)/3;
            if(area(stk[r],pr,stk[ll])<=area(stk[r],pr,stk[rr]))L=ll;else R=rr;
        }
         
        if(getdis(stk[r],pr)*(h1+h2)<res){
            res=getdis(stk[r],pr)*(h1+h2);
            re[0]=move(pr,stk[i]-stk[i+1],h1);
            re[1]=move(stk[r],stk[i]-stk[i+1],h1);
            re[2]=move(stk[r],stk[i+1]-stk[i],h2);
            re[3]=move(pr,stk[i+1]-stk[i],h2);
        }
    }
     
    printf("%.5lf\n",res);
    int bins;Point Minre;
    for(Minre=re[0],bins=0,i=1;i<4;++i)if(dcmp(re[i].y-Minre.y)<0||(dcmp(re[i].y-Minre.y)==0&&dcmp(re[i].x-Minre.x)<0))Minre=re[i],bins=i;
     
    for(i=0;i<4;++i)re[(i+bins)%4].print();
     
    return 0;
}

思路:

显然一条比较优的路径是从\(A\)走到\(AB\)上一点\(X\),然后从\(X\)走到\(CD\)上一点\(Y\),然后从\(Y\)走到\(D\).

关键是\(X,Y\)如何确定.

我们能够发现\(X\)点确定时,随着\(Y\)的移动,总路程是一个单峰函数.

更加神奇的是,随着\(X\)的移动,通过我们上述三分得到的总路程也是一个单峰函数!

没什么说的,三分套三分就行了.

#include<cstdio>
#include<cstring>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
 
typedef double f2;
 
#define eps 1e-8
struct Point{
    f2 x,y;
    void read(){scanf("%lf%lf",&x,&y);}
    Point(){}
    Point(f2 _x,f2 _y):x(_x),y(_y){}
    Point operator-(const Point&B)const{return Point(B.x-x,B.y-y);}
    Point operator+(const Point&B)const{return Point(x+B.x,y+B.y);}
    Point operator*(const f2&p)const{return Point(p*x,p*y);}
}A,B,C,D;
 
int vab,vcd,vplane;
 
f2 sqr(const f2&x){return x*x;}
f2 Dist(const Point&A,const Point&B){return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));}
f2 f(Point x,Point l,Point v,f2 k){
    return Dist(x,l+v*k)/vplane+Dist(l+v*k,l+v)/vcd;
}
f2 Solve(Point x,Point l,Point r){
    Point v=l-r;
    f2 L=0,R=1,LL,RR,ansl,ansr,lastans=1e12,nowans;
    while(1){
        LL=L+(R-L)/3,RR=R-(R-L)/3;
        ansl=f(x,l,v,LL),ansr=f(x,l,v,RR);
        if(ansl<ansr)R=RR,nowans=ansl;else L=LL,nowans=ansr;
        if(fabs(nowans-lastans)<1e-5)break;lastans=nowans;
    }
    return lastans;
}
 
int main(){
    A.read(),B.read(),C.read(),D.read();
    scanf("%d%d%d",&vab,&vcd,&vplane);
    f2 L=0,R=1,LL,RR,ansl,ansr,lastans=1e12,nowans;Point v=A-B;
    while(1){
        LL=L+(R-L)/3,RR=R-(R-L)/3;
        ansl=Dist(A,A+v*LL)/vab+Solve(A+v*LL,C,D),ansr=Dist(A,A+v*RR)/vab+Solve(A+v*RR,C,D);
        if(ansl<ansr)R=RR,nowans=ansl;else L=LL,nowans=ansr;
        if(fabs(nowans-lastans)<1e-5)break;lastans=nowans;
    }
    printf("%.2lf",lastans);
    return 0;
}