bzoj1038(半平面交)

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=;
const double eps=1e-;
double ans=1e10;
int n,head,tail;
struct vec{
    double x,y;
    vec(double x=,double y=):x(x),y(y){}
    vec operator-(vec& a){
        return vec(x-a.x,y-a.y);
    }
    vec operator+(vec&a){
        return vec(x+a.x,y+a.y);
    }
}po[maxn],g[maxn];
vec operator*(vec a,double t){return vec(a.x*t,a.y*t);}
double cross(vec a,vec b){return a.x*b.y-b.x*a.y;}
struct lin{
    vec p,v;//p是直线上一个点，v是方向向量
    double ang;//ang是斜率;
    lin(){}
    lin(vec p,vec v):p(p),v(v){ang=atan2(v.y,v.x);}
    bool operator<(const lin&a)const{
        return ang<a.ang;
    }
}ll[maxn],q[maxn],bb,cc;
bool onl(lin L,vec p){
    return cross(L.v,p-L.p)>;
}
vec qj(lin a,lin b){
    vec u=a.p-b.p;
    double t=cross(b.v,u)/cross(a.v,b.v);
    return a.v*t+a.p;
}
void halfj(){
    sort(ll,ll+n);
    q[head=tail=]=ll[];
    for(int i=;i<n;++i){
        while(head<tail&&!onl(ll[i],g[tail-]))tail--;
        while(head<tail&&!onl(ll[i],g[head]))head++;
        q[++tail]=ll[i];
        if(fabs(cross(q[tail].v,q[tail-].v))<eps){
            --tail;if(onl(q[tail],ll[i].p))q[tail]=ll[i];
        }
        if(head<tail)g[tail-]=qj(q[tail-],q[tail]);
    }
    while(head<tail&&!onl(q[head],g[tail-]))--tail;
}
int main(){
    cin>>n;
    for(int i=;i<=n;++i){
        scanf("%lf",&po[i].x);
    }
    for(int i=;i<=n;++i){
        scanf("%lf",&po[i].y);
    }
    po[].x=po[].x;po[].y=;
    po[n+].x=po[n].x;po[n+].y=;
    for(int i=;i<=n;++i){
        ll[i]=lin(po[i],po[i+]-po[i]);
    }
    n+=;
    halfj();
    for(int k=;k<tail;++k)
        for(int i=;i<n-;++i){
            vec tmp;tmp.x=g[k].x;tmp.y=-;
            if(g[k].x>=po[i].x&&g[k].x<=po[i+].x){
                cc=lin(po[i],po[i+]-po[i]);
                bb=lin(tmp,g[k]-tmp);
                ans=min(ans,g[k].y-qj(cc,bb).y);
            }
    }
    for(int k=;k<=n-;++k)
        for(int i=;i<tail-;++i){
            vec tmp;tmp.x=po[k].x;tmp.y=-;
            if(po[k].x>=g[i].x&&po[k].x<=g[i+].x){
                cc=lin(g[i],g[i+]-g[i]);
                bb=lin(tmp,po[k]-tmp);
                ans=min(ans,qj(cc,bb).y-po[k].y);
            }
    }
    printf("%.3lf",ans);
    return ;
}