思路:

首先我们将所有的点都顺时针旋转\(a\)度,这样就相当于坐标系逆时针旋转了\(a\)度,那么现在椭圆的长轴就与\(x\)轴平行了.

那么还要考虑椭圆现在是横向伸长了\(p\)倍,于是我们就将所有点目前的横坐标缩小\(p\)倍.

然后就是随机增量法的最小圆覆盖了.

#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);}
     
    bool operator<(const Point&B)const{return dcmp(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(p*x,p*y);}
    Point operator/(const f2&p)const{return Point(x/p,y/p);}
}P[N];
 
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;}
 
Point getmid(const Point&a,const Point&b){return Point((a.x+b.x)/2,(a.y+b.y)/2);}
Point getvec(const Point&v){return Point(-v.y,v.x);}
Point rot(const Point&p,f2 alp){return Point(p.x*cos(alp)-p.y*sin(alp),p.x*sin(alp)+p.y*cos(alp));}
 
struct Line{
    Point p,v;
    Line(){}
    Line(const Point&p1,const Point&p2):p(p1),v(p1-p2){}
};
Line getvecline(const Point&p1,const Point&p2){
    Line re;
    re.p=getmid(p1,p2),re.v=getvec(p1-p2);
    return re;
}
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 getpoint(Point p1,Point p2,Point p3){
    static Point P[3];P[0]=p1,P[1]=p2,P[2]=p3;sort(P,P+3);p1=P[0],p2=P[1],p3=P[2];
    if(dcmp(cross(p1-p2,p2-p3))==0)return getmid(p1,p3);
    else return getlineintersection(getvecline(p1,p2),getvecline(p2,p3));
}
 
int main(){
    int n;scanf("%d",&n);register int i,j,k;
    for(i=1;i<=n;++i)P[i].read();
     
    f2 alp,p;scanf("%lf%lf",&alp,&p);
     
    for(i=1;i<=n;++i)P[i]=rot(P[i],-alp/180*acos(-1.0)),P[i].x/=p;
     
    for(i=2;i<=n;++i)swap(P[i],P[rand()%i+1]);
    Point o=P[1];f2 r=0;
    for(i=2;i<=n;++i)if(getdis(o,P[i])>r){
        o=P[i],r=0;
        for(j=1;j<i;++j)if(getdis(o,P[j])>r){
            o=getmid(P[i],P[j]),r=getdis(o,P[i]);
            for(k=1;k<j;++k)if(getdis(o,P[k])>r)o=getpoint(P[i],P[j],P[k]),r=getdis(o,P[i]);
        }
    }
     
    printf("%.3lf",r);
     
    return 0;
}

思路:

我们发现最终的答案就是缩小一圈后的矩形的顶点们形成的凸包的周长再加上一个半径为\(r\)的圆的周长.

因为多边形外角和为\(360\)度.这个不用多说.

此外还有一点点问题就是旋转后的坐标变换.

如果是\((x,y)\)以原点为旋转中心逆时针旋转\(\alpha\)弧度,那么新的坐标为\((xcos\alpha-ysin\alpha,xsin\alpha+ycos\alpha)\).

#include<cstdio>
#include<cstring>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
 
typedef double f2;
 
#define N 200010
 
static const f2 eps=1e-8;
inline int dcmp(const f2&x){return (fabs(x)<eps)?0:(x<0?-1:1);}
 
struct Point{
    f2 x,y;
    Point(){}
    Point(f2 _x,f2 _y):x(_x),y(_y){}
     
    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(p*x,p*y);}
    Point operator/(const f2&p)const{return Point(x/p,y/p);}
     
    bool operator<(const Point&B)const{return dcmp(x-B.x)==0?(dcmp(y-B.y)<0):(dcmp(x-B.x)<0);}
}P[N<<2],PP[N<<2],stk[N<<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));
}
Point move(const Point&p,const f2&Ang,const f2&x,const f2&y){
    return Point(p.x*cos(Ang)-p.y*sin(Ang)+x,p.x*sin(Ang)+p.y*cos(Ang)+y);
}
 
int main(){
    int n;scanf("%d",&n);
     
    f2 a,b,r,x,y,Ang;scanf("%lf%lf%lf",&a,&b,&r);a-=2*r,b-=2*r;
     
    register int i,j;
    int cnt=0;
    for(i=1;i<=n;++i){
        scanf("%lf%lf%lf",&x,&y,&Ang);
         
        P[++cnt]=move(Point(b/2,a/2),Ang,x,y);
        P[++cnt]=move(Point(-b/2,a/2),Ang,x,y);
        P[++cnt]=move(Point(b/2,-a/2),Ang,x,y);
        P[++cnt]=move(Point(-b/2,-a/2),Ang,x,y);
    }
     
    int top=0;
    sort(P+1,P+cnt+1);
    int revins;for(revins=cnt;revins!=1&&P[revins].x==P[revins-1].x;--revins);
    for(i=revins;i<=(n+revins)/2;++i)swap(P[i],P[n+revins-i]);
     
    for(i=1;i<=cnt;++i){
        if(!top)stk[++top]=P[i];
        else{while(top>1&&(P[i].y-stk[top].y)*(stk[top].x-stk[top-1].x)<=(stk[top].y-stk[top-1].y)*(P[i].x-stk[top].x))--top;stk[++top]=P[i];}
    }
    int nowtop=top;
    for(i=revins-1;i>=1;--i){
        if(top==nowtop)stk[++top]=P[i];
        else{while(top>=nowtop+1&&(P[i].y-stk[top].y)*(stk[top].x-stk[top-1].x)<=(stk[top].y-stk[top-1].y)*(P[i].x-stk[top].x))--top;stk[++top]=P[i];}
    }--top;
     
    f2 re=0;
    for(i=1;i<top;++i)re+=getdis(stk[i],stk[i+1]);re+=getdis(stk[top],stk[1]);
     
    printf("%.2lf",re+2*acos(-1.0)*r);
     
    return 0;
}