题意给的是一个斜着的的椭圆形,它的长轴距离x轴的正方向有a度
然而我们并不会求最小椭圆覆盖
因此我们处理点。
把所有点先向右旋转a度,然后再缩小P倍数
然后求最小圆的覆盖就行
#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
typedef long long ll;
const double eps=1e-10;
const double pi=acos(-1);
int sgn(double x)
{
if (fabs(x)<eps) return 0;
return x<0?-1:1;
}
struct Point
{
double x,y;
Point(){}
Point (double x,double y):x(x),y(y) {}
Point operator /(double k){return Point(x/k,y/k);}
};
const int N=1e5+100;
Point p[N];
double Distance(Point A,Point B)
{
return hypot(A.x-B.x,A.y-B.y);
}
Point circle_center(const Point a,const Point b,const Point c)
{
Point center;
double a1=b.x-a.x,b1=b.y-a.y,c1=(a1*a1+b1*b1)/2;
double a2=c.x-a.x,b2=c.y-a.y,c2=(a2*a2+b2*b2)/2;
double d=a1*b2-a2*b1;
center.x=a.x+(c1*b2-c2*b1)/d;
center.y=a.y+(a1*c2-a2*c1)/d;
return center;
}
void min_cover_circle(Point *p,int n,Point &c,double &r)
{
random_shuffle(p,p+n);
c=p[0],r=0;
for(int i=1; i<n; i++)
{
if (sgn(Distance(p[i],c)-r)>0)
{
c=p[i],r=0;
for(int j=0; j<i; j++)
{
if (sgn(Distance(p[j],c)-r)>0)
{
c.x=(p[i].x+p[j].x)/2;
c.y=(p[i].y+p[j].y)/2;
r=Distance(p[j],c);
for(int k=0; k<j; k++)
{
if (sgn(Distance(p[k],c)-r)>0)
{
c=circle_center(p[i],p[j],p[k]);
r=Distance(p[i],c);
}
}
}
}
}
}
}
Point Rotate(Point A,double rad)
{
return Point(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
int main()
{
int n;
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
double rad,pp;
scanf("%lf%lf",&rad,&pp);
rad=rad/180.0*pi;
for(int i=0;i<n;i++)
{
p[i]=Rotate(p[i],-rad);
p[i].x=p[i].x/pp;
}
double r;
Point c;
min_cover_circle(p,n,c,r);
printf("%.3lf",r);
return 0;
}
考虑过墙外调查吗
塔塔开
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