bzoj 1038: [ZJOI2008]瞭望塔

题意大概就是在村子（山）轮廓之上建一个多么高的塔，可以看到村子全部的点。（一开始怎么也看不懂题意2333）

显然，相邻两点连线，然后所有线交出的区域就是塔建造的可行范围，所以就可以半平面交了233。

答案最小，，，就在分段函数的分段端点处取？？？（雾，画个图能看出来）

如此神的题%%hzwer

#include<cstdio>
#include<algorithm>
#include<cmath>
#define eps 1e-8
using namespace std;
double ans=1e60;
int n,cnt,top,tot;
struct point{double x,y;}p[1005],a[1005];
struct line{point a,b; double angle;} l[1005],q[1005];
point operator - (point a, point b){
    point t; t.x=a.x-b.x; t.y=a.y-b.y; return t;
}
double operator * (point a, point b){
    return a.x*b.y-a.y*b.x;
}
bool operator < (line a, line b){
    if (a.angle==b.angle) return (a.b-a.a)*(b.b-a.a)>0;
    return a.angle<b.angle;
}
point intersection(line a, line b)
{
    double k1,k2,t;
    k1=(b.b-a.a)*(a.b-a.a);
    k2=(a.b-a.a)*(b.a-a.a);
    t=k1/(k1+k2);
    point ans;
    ans.x=b.b.x+(b.a.x-b.b.x)*t;
    ans.y=b.b.y+(b.a.y-b.b.y)*t;
    return ans;
}
bool jud(line a, line b, line t)
{
    point p=intersection(a,b);
    return (t.b-t.a)*(p-t.a)<0;
}
void half_plane_intersection()
{
    int L=1,R=0; tot=0;
    for (int i=1; i<=cnt; i++)
    {
        if (l[i].angle!=l[i-1].angle) tot++;
        l[tot]=l[i];
    }
    cnt=tot;
    q[++R]=l[1]; q[++R]=l[2];
    for (int i=3; i<=cnt; i++)
    {
        while (L<R && jud(q[R-1],q[R],l[i])) R--;
        while (L<R && jud(q[L+1],q[L],l[i])) L++;
        q[++R]=l[i]; 
    }
    while (L<R && jud(q[R-1],q[R],q[L])) R--;
    while (L<R && jud(q[L+1],q[L],q[R])) L++;
    tot=0;
    for (int i=L; i<R; i++)
        a[++tot]=intersection(q[i],q[i+1]);
}
void pre()
{
    p[0].x=p[1].x; p[0].y=100001;
    p[n+1].x=p[n].x; p[n+1].y=100001;
    for (int i=1; i<=n; i++)
    {
        l[++cnt].a=p[i-1]; l[cnt].b=p[i];
        l[++cnt].a=p[i]; l[cnt].b=p[i+1];
    }
    for (int i=1; i<=cnt; i++)
        l[i].angle=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x);
    sort(l+1,l+cnt+1);
}
void get_ans()
{
    for (int k=1; k<=tot; k++)
        for (int i=1; i<n; i++)
        {
            point t; t.x=a[k].x; t.y=-1;
            if (a[k].x>=p[i].x && a[k].x<=p[i+1].x)
                ans=min(ans,a[k].y-intersection((line){p[i],p[i+1]},(line){t,a[k]}).y);
        }
    for (int k=1; k<=n; k++)
        for (int i=1; i<tot; i++)
        {
            point t; t.x=p[k].x; t.y=-1;
            if (p[k].x>=a[i].x && p[k].x<=a[i+1].x)
                ans=min(ans,intersection((line){a[i],a[i+1]},(line){t,p[k]}).y-p[k].y);
        }
        
}
int main()
{
    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);
    pre();
    half_plane_intersection();
    get_ans();
    printf("%.3lf",ans);
    return 0;
}