BZOJ 1038 半平面交

多么明显的半平面交。

答案一定在山的顶点处或者半平面交区域的顶点处。。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
 
#define N 555
#define EPS 1e-7
#define INF 1e12
 
using namespace std;
 
struct PO
{
    double x,y;
    void prt() {printf("%lf     %lf\n",x,y);}
}p[N],tp[N],s[N];
 
int n,m;
 
inline PO operator +(PO a,PO b)
{
    a.x+=b.x; a.y+=b.y;
    return a;
}
 
inline PO operator -(PO a,PO b)
{
    a.x-=b.x; a.y-=b.y;
    return a;
}
 
inline PO operator *(PO a,double k)
{
    a.x*=k; a.y*=k;
    return a;
}
 
inline PO operator /(PO a,double k)
{
    a.x/=k; a.y/=k;
    return a;
}
 
inline int dc(double x)
{
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}
 
inline double cross(const PO &a,const PO &b,const PO &c)
{
    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
 
inline double dot(const PO &a,const PO &b,const PO &c)
{
    return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);
}
 
inline PO getpoint(const PO &a,const PO &b,const PO &c,const PO &d)
{
    double k1=cross(a,d,c),k2=cross(b,c,d);
    return a+(b-a)*k1/(k1+k2);
}
 
inline bool onseg(const PO &a,const PO &b,const PO &c)
{
    if(dc(dot(a,b,c))<=0) return true;
    return false;
}
 
inline void read()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lf",&p[i].x);
    for(int i=1;i<=n;i++) scanf("%lf",&p[i].y);
}
 
inline void getcut()
{
    tp[1].x=tp[5].x=-INF; tp[1].y=tp[5].y=-INF;
    tp[2].x=INF,tp[2].y=-INF;
    tp[3].x=INF,tp[3].y=INF;
    tp[4].x=-INF,tp[4].y=INF;
    int cp=4,tc;
    for(int i=1;i<n;i++)
    {
        tc=0;
        for(int j=1;j<=cp;j++)
        {
            if(dc(cross(p[i],p[i+1],tp[j]))>=0) s[++tc]=tp[j];
            if(dc(cross(p[i],p[i+1],tp[j])*cross(p[i],p[i+1],tp[j+1]))<0)
                s[++tc]=getpoint(p[i],p[i+1],tp[j],tp[j+1]);
        }
        s[tc+1]=s[1];
        for(int j=1;j<=tc+1;j++) tp[j]=s[j];
        cp=tc;
    }
    m=cp;
}
 
inline void go()
{
    getcut();
    double ans=INF;
    PO s1,t1,jd;
    for(int i=1;i<=n;i++)
    {
        s1=p[i]; t1.x=s1.x,t1.y=520.1314;
        for(int j=1;j<=m;j++)
        {
            jd=getpoint(s1,t1,s[j],s[j+1]);
            if(onseg(jd,s[j],s[j+1])) ans=min(ans,jd.y-p[i].y);
        }
    }
    for(int i=1;i<=m;i++)
    {
        s1=s[i]; t1.x=s[i].x; t1.y=520.1314;
        for(int j=1;j<n;j++)
        {
            jd=getpoint(s1,t1,p[j],p[j+1]);
            if(onseg(jd,p[j],p[j+1])) ans=min(ans,s[i].y-jd.y);
        }
    }
    printf("%.3lf\n",ans);
}
 
int main()
{
    read(),go();
    return 0;
}