把地面看成半平面，能看到所有位置的点所在的区域即为半平面的交

因为分段函数的极值只会在转折处或边界取到，所以对于半平面上和地面上的每一个交点都求一下距离就好了

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e3+100;
const double maxx=4e12;
struct Point{
    double x,y;
    Point(double xx=0,double yy=0){
        x=xx,y=yy;
    }
}a[maxn],p[maxn];
struct Vector{
    double x,y;
    Vector(double xx=0,double yy=0){
        x=xx,y=yy;
    }
};
struct Line{
    Point p;
    Vector v;
    double ang;
    Line(Point a=Point(),Vector b=Vector()){
        p=a,v=b;
        ang=atan2(v.y,v.x);
    }
}q[maxn],b[maxn];
int dcmp(double x){return fabs(x)<1e-10?0:(x>0?1:-1);}
Vector operator - (Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}
Point operator + (Point a,Vector b){return Point(a.x+b.x,a.y+b.y);}
Vector operator * (double p,Vector a){return Vector(a.x*p,a.y*p);}
double operator * (Vector a,Vector b){return a.x*b.y-a.y*b.x;}
double operator * (Point a,Point b){return a.x*b.y-a.y*b.x;}
bool operator < (Point a,Point b){return dcmp(a.x-b.x)==0?(dcmp(a.y-b.y)<0):(a.x<b.x);}
bool operator < (Line a,Line b){return dcmp(a.ang-b.ang)==0?(dcmp(a.v*(b.p-a.p))>0):(a.ang<b.ang);}
double len(Vector a){return sqrt(a.x*a.x+a.y*a.y);}
Point glt(Line a,Line b){Vector v=a.p-b.p; return a.p+b.v*v/(a.v*b.v)*a.v;}
bool onright(Line a,Line b,Line t){Point p=glt(a,b); return dcmp(t.v*(p-t.p))<0;}
void bpm(Line *b,int &n,Point *p){
    sort(b+1,b+n+1);
    int l=0,r=1,tot=0;
    for(int i=1;i<=n;i++){
        if(b[i].ang!=b[i-1].ang) tot++;
        b[tot]=b[i];
    }
    n=tot,q[0]=b[1],q[1]=b[2];
    for(int i=3;i<=n;i++){
        while(l<r&&onright(q[r],q[r-1],b[i])) r--;
        while(l<r&&onright(q[l],q[l+1],b[i])) l++;
        q[++r]=b[i];
    }
    while(l<r&&onright(q[r],q[r-1],q[l])) r--;
    while(l<r&&onright(q[l],q[l+1],q[r])) l++;
    n=0,q[r+1]=q[l];
    for(int i=l;i<=r;i++)
        b[++n]=q[i],p[n]=glt(q[i],q[i+1]);
}
int n;
void solve(Point *a,int m,Point *p,int n){
    sort(p+1,p+n+1);
//    for(int i=1;i<=n;i++) printf("%lf %lf\n",p[i].x,p[i].y);
    int l=2;
    double ans=maxx*2;
    for(int i=1;i<n;i++){
        while(l<m&&dcmp(p[i].x-a[l+1].x)>0) l++;
        Point pz=glt(Line(a[l],a[l+1]-a[l]),Line(p[i],Vector(0,1)));
        ans=min(ans,len(pz-p[i]));
    }
    l=1;
    for(int i=2;i<m;i++){
        while(l<n&&dcmp(a[i].x-p[l+1].x)>0) l++;
        Point pz=glt(Line(p[l],p[l+1]-p[l]),Line(a[i],Vector(0,1)));
        ans=min(ans,len(pz-a[i]));
    }
    printf("%.3lf\n",ans);
}
int main(){
    scanf("%d",&n),n++;
    for(int i=2;i<=n;i++) scanf("%lf",&a[i].x);
    for(int i=2;i<=n;i++) scanf("%lf",&a[i].y);
    a[1]=Point(a[2].x,maxx),a[n+1]=Point(a[n].x,maxx);
    int m=n+1;
    for(int i=1;i<=n;i++)
        b[i]=Line(a[i],a[i+1]-a[i]);
    bpm(b,n,p);
    solve(a,m,p,n);
    return 0;
}