題目鏈接：http://www.lydsy.com/JudgeOnline/problem.php?id=3564

題意：給出平面上n個(gè)點(diǎn)，畫出一個(gè)橢圓，橢圓的長軸是短軸的p倍，且長軸的方向?yàn)閤軸逆時(shí)針旋轉(zhuǎn)a度。求這個(gè)橢圓短軸的最小值使得可以覆蓋所以點(diǎn)。

思路：先將所有點(diǎn)順時(shí)針旋轉(zhuǎn)a，然后所有點(diǎn)的x縮為原來的1/p。然后就是最小圓覆蓋。

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const int N=50005;struct point {double x,y;point(double _x=0,double _y=0){x=_x;y=_y;}point operator-(point a){return point(x-a.x,y-a.y);}point operator+(point a){return point(x+a.x,y+a.y);}double operator*(point a){return x*a.y-y*a.x;}point operator*(double t){return point(x*t,y*t);}point operator/(double t){return point(x/t,y/t);}point zhuan(double ang){return point(x*cos(ang)+y*sin(ang),x*sin(ang)-y*cos(ang));}point verl(){return point(-y,x);} };point p[N]; int n;double dis(point a,point b) {return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y)); }int sgn(double x) {if(x>1e-20) return 1;if(x<-1e-20) return -1;return 0; }point cross(point a,point b,point p,point q) {double s1=(q-a)*(p-a);double s2=(p-b)*(q-b);return (a*s2+b*s1)/(s1+s2); }point get(point a,point b,point c) {double x=(b-a)*(c-a);if(sgn(x)==0){double ab=dis(a,b);double bc=dis(b,c);double ac=dis(a,c);if(sgn(ab-bc)>=0&&sgn(ab-ac)>=0) return (a+b)/2;if(sgn(bc-ab)>=0&&sgn(bc-ac)>=0) return (b+c)/2;return (a+c)/2;}point M1=(a+b)/2,v1=(a-b).verl();point M2=(b+c)/2,v2=(b-c).verl();return cross(M1,M1+v1,M2,M2+v2); }int main() {scanf("%d",&n);int i;for(i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);double ang,scal;scanf("%lf%lf",&ang,&scal);ang=ang/180*PI;for(i=0;i<n;i++){p[i]=p[i].zhuan(ang);p[i].x/=scal;}double r=0;point c=p[0];for(i=1;i<n;i++) if(dis(p[i],c)>r+1e-10){c=p[i]; r=0;int j;for(j=0;j<i;j++) if(dis(p[j],c)>r+1e-10){c=(p[i]+p[j])/2;r=dis(p[i],p[j])/2;int k;for(k=0;k<j;k++) if(dis(p[k],c)>r+1e-10){c=get(p[i],p[j],p[k]);r=max(dis(p[i],c),max(dis(p[k],c),dis(p[j],c)));}}}printf("%.3lf\n",r); }

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