BZOJ 3564 信号增幅仪

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3564

题意：给出平面上n个点，画出一个椭圆，椭圆的长轴是短轴的p倍，且长轴的方向为x轴逆时针旋转a度。求这个椭圆短轴的最小值使得可以覆盖所以点。

思路：先将所有点顺时针旋转a，然后所有点的x缩为原来的1/p。然后就是最小圆覆盖。

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);
}