[SHOI2014]信号增幅仪

题目大意：
　　平面直角坐标系中散落着n个点，一个椭圆的长半轴在对于x轴逆时针旋转α度的角度上，且长半轴是短半轴的k倍。
　　问短半轴至少要多长才能覆盖所有的点？

思路：
　　首先把坐标顺时针旋转α度，然后把所有点的横坐标缩小k倍，就变成了最小圆覆盖问题。

#include<cmath>
#include<cstdio>
#include<cctype>
#include<algorithm>
inline int getint() {
    register char ch;
    register bool neg=false;
    while(!isdigit(ch=getchar())) if(ch=='-') neg=true;
    register int x=ch^'0';
    while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
    return neg?-x:x;
}
const int N=50000;
const double eps=1e-4;
struct Point {
    double x,y;
};
Point p[N];
inline double sqr(const double &x) {
    return x*x;
}
inline double dis(const Point &a,const Point &b) {
    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
}
inline Point mid(const Point &a,const Point &b) {
    return (Point){(a.x+b.x)/2,(a.y+b.y)/2};
}
inline Point out(const Point &a,const Point &b,const Point &c) {
    Point ret;
    ret.x=((sqr(a.x)+sqr(a.y))*b.y+(sqr(c.x)+sqr(c.y))*a.y+(sqr(b.x)+sqr(b.y))*c.y-(sqr(a.x)+sqr(a.y))*c.y-(sqr(c.x)+sqr(c.y))*b.y-(sqr(b.x)+sqr(b.y))*a.y)/(a.x*b.y+b.x*c.y+c.x*a.y-a.x*c.y-b.x*a.y-c.x*b.y)/2;
    ret.y=((sqr(a.x)+sqr(a.y))*c.x+(sqr(c.x)+sqr(c.y))*b.x+(sqr(b.x)+sqr(b.y))*a.x-(sqr(a.x)+sqr(a.y))*b.x-(sqr(c.x)+sqr(c.y))*a.x-(sqr(b.x)+sqr(b.y))*c.x)/(a.x*b.y+b.x*c.y+c.x*a.y-a.x*c.y-b.x*a.y-c.x*b.y)/2;
    return ret;
}
int main() {
    const int n=getint();
    for(register int i=0;i<n;i++) {
        p[i]=(Point){getint(),getint()};
    }
    const double alpha=getint()*M_PI/180,k=getint();
    for(register int i=0;i<n;i++) {
        const double x=p[i].x,y=p[i].y;
        p[i]=(Point){(x*cos(alpha)+y*sin(alpha))/k,y*cos(alpha)-x*sin(alpha)};
    }
    std::random_shuffle(&p[0],&p[n]);
    Point c=p[0];
    double r=0;
    for(register int i=1;i<n;i++) {
        if(dis(c,p[i])<r+eps) continue;
        c=p[i];
        r=0;
        for(register int j=0;j<i;j++) {
            if(dis(c,p[j])<r+eps) continue;
            c=mid(p[i],p[j]);
            r=dis(c,p[j]);
            for(register int k=0;k<j;k++) {
                if(dis(c,p[k])<r+eps) continue;
                c=out(p[i],p[j],p[k]);
                r=dis(c,p[k]);
            }
        }
    }
    printf("%.3f\n",r);
    return 0;
}