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题解

我连椭圆是个啥都不知道导致这么简单一道题我一点思路都没有……

我们把坐标系旋转一下，让半长轴成为新的\(x\)轴，也就是说所有点都绕原点逆时针旋转\(360-a\)度，然后再把所有点的\(x\)坐标变为原来的\({1\over p}\)，跑一个最小圆覆盖就行了

//minamoto

#include<bits/stdc++.h>

#define R register

#define inline __inline__ __attribute__((always_inline))

#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)

#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)

#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)

using namespace std;

char buf[1<<21],*p1=buf,*p2=buf;

inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}

int read(){

    R int res,f=1;R char ch;

    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);

    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');

    return res*f;

}

double readdb()

{

    R double x=0,y=0.1,f=1;R char ch;

    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);

    for(x=ch-'0';(ch=getc())>='0'&&ch<='9';x=x*10+ch-'0');

    for(ch=='.'&&(ch=getc());ch>='0'&&ch<='9';x+=(ch-'0')*y,y*=0.1,ch=getc());

    return x*f;

}

const int N=1e5+5;const double eps=1e-6,Pi=acos(-1.0);

inline double reps(){return (1.0*rand()/RAND_MAX-0.5)*eps;}

inline int sgn(R double x){return x<-eps?-1:x>eps;}

struct node{

    double x,y;

    inline node(){}

    inline node(R double xx,R double yy):x(xx),y(yy){}

    inline node operator +(const node &b)const{return node(x+b.x,y+b.y);}

    inline node operator -(const node &b)const{return node(x-b.x,y-b.y);}

    inline double operator *(const node &b)const{return x*b.y-y*b.x;}

    inline node operator *(const double &b)const{return node(x*b,y*b);}

    inline double operator ^(const node &b)const{return x*b.x+y*b.y;}

    inline double len2(){return x*x+y*y;}

    inline node rot(R double s,R double c){return node(x*c-y*s,x*s+y*c);}

    inline node rot90(){return node(-y,x);}

    inline void shake(){x+=reps(),y+=reps();}

}p[N],o;

struct Line{

    node p,v;

    inline Line(){}

    inline Line(R node pp,R node vv):p(pp),v(vv){}

    friend node cross(const Line &a,const Line &b){return a.p+a.v*(b.v*(b.p-a.p)/(b.v*a.v));}

};

node circle(const node &a,const node &b,const node &c){

    return cross(Line((a+b)*0.5,(b-a).rot90()),Line((a+c)*0.5,(c-a).rot90()));

}

int n;double s,c,a,b;

double calc(){

	double r=0;o=node(0,0);

	random_shuffle(p+1,p+1+n);

    fp(i,1,n)if(sgn((p[i]-o).len2()-r)>0){

        o=p[i],r=0;

        fp(j,1,i-1)if(sgn((p[j]-o).len2()-r)>0){

            o=(p[i]+p[j])*0.5,r=(p[j]-o).len2();

            fp(k,1,j-1)if(sgn((p[k]-o).len2()-r)>0)

                o=circle(p[i],p[j],p[k]),r=(p[k]-o).len2();

        }

    }

    return sqrt(r);

}

int main(){

    srand(20030719);

//	freopen("testdata.in","r",stdin);

    n=read();

    fp(i,1,n)p[i].x=readdb(),p[i].y=readdb(),p[i].shake();

    b=read(),a=read(),b=(360-b)/180.0*Pi,s=sin(b),c=cos(b),a=1.0/a;

    fp(i,1,n)p[i]=p[i].rot(s,c);

    fp(i,1,n)p[i].x=p[i].x*a;

    printf("%.3lf\n",calc());

    return 0;

}