[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^'';
     while(isdigit(ch=getchar())) x=(((x<<)+x)<<)+(ch^'');
     return neg?-x:x;
 }
 const int N=;
 const double eps=1e-;
 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)/,(a.y+b.y)/};
 }
 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)/;
     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)/;
     return ret;
 }
 int main() {
     const int n=getint();
     for(register int i=;i<n;i++) {
         p[i]=(Point){getint(),getint()};
     }
     const double alpha=getint()*M_PI/,k=getint();
     for(register int i=;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[],&p[n]);
     Point c=p[];
     double r=;
     for(register int i=;i<n;i++) {
         if(dis(c,p[i])<r+eps) continue;
         c=p[i];
         r=;
         for(register int j=;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=;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 ;
 }