ZJOI2008 瞭望塔 半平面交

题意：给出一个以n个点为轮廓的村庄，在村庄任意位置放一个瞭望塔，使瞭望塔能看到村庄的所有位置，求瞭望塔最低高度。

只我们发现只有这个点在每个直线所在半平面以上的时候才能看到，如样例图:

还注意到，只有在原图的端点或半平面交的端点处才会更新答案。

#include
#include
#include
#include
using namespace std;
const double esp=1e-8;
const int maxn=310;
int n,m;
int dcmp(double x){ return x>esp?1:(x<-esp?-1:0); }
struct Point{
    double x,y;
    Point(){}
    Point(double X,double Y):x(X),y(Y){}
    bool operator <(const Point& A)const{
        return xx;
    }
};
typedef Point Vector;
Vector operator -(const Point& A,const Point& B){ return Vector(A.x-B.x,A.y-B.y); }
Point operator +(const Point& A,const Vector& B){ return Vector(A.x+B.x,A.y+B.y); }
Vector operator *(const Vector& A,const double& B){ return Vector(A.x*B,A.y*B); }
double Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x; }
struct Line{
    Point P;
    Vector v;
    double ang;
    Line(){}
    Line(const Point& A,const Point& B){
        v=B-A;
        P=A;
        ang=atan2(v.y,v.x);
    }
    bool operator <(const Line& A)const{
        return angreturn Cross(L.v,P-L.P)>0;
}
Point p[maxn],land[maxn];
Line q[maxn],L[maxn];
Point GetIntersection(Line a,Line b){
    Vector u=a.P-b.P;
    double t=Cross(b.v,u)/Cross(a.v,b.v);
    return a.P+a.v*t;
}
void HalfPlaneIntersection(){
    int l=1,r=1;
    sort(L+1,L+1+n);
    q[l]=L[1];
    for(int i=2;i<=n;i++){
        while(l1])) r--;
        while(lq[++r]=L[i];
        if(lq[r].v,q[r-1].v))){
            r--;
            if(Onleft(q[r],L[i].P)) q[r]=L[i];
        }
        if(l1]=GetIntersection(q[r-1],q[r]);
    }
    while(lq[l],p[r-1])) r--;
    p[r]=GetIntersection(q[r],q[l]);
    for(int i=l;i<=r;i++) q[i-l+1]=q[i];
    m=r-l+1;
}
int x[maxn],y[maxn];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&x[i]);
    for(int i=1;i<=n;i++) scanf("%d",&y[i]),land[i]=Point(x[i],y[i]);
    for(int i=1;i1]);
    L[n]=(Line){Point(1e12,1e12),Point(-1e12,1e12)};
    HalfPlaneIntersection();
    sort(p+1,p+1+m);
    double ans=1e12;
    for(int i=1;i<=n;i++)
        for(int j=1;j<m;j++){
            if(dcmp(x[i]-p[j].x)>=0 && dcmp(x[i]-p[j+1].x)<=0){
                ans=min(ans,p[j].y+(p[j+1].y-p[j].y)/(p[j+1].x-p[j].x)*(x[i]-p[j].x)-y[i]);
                break;
            }
        }
    for(int j=1;j<=m;j++)
        for(int i=1;iif(dcmp(p[j].x-x[i])>=0 && dcmp(p[j].x-x[i+1])<=0){
                ans=min(ans,p[j].y-(1.0*y[i]+1.0*(y[i+1]-y[i])/(x[i+1]-x[i])*(p[j].x-x[i])));
                break;
            }
        }
    printf("%.3lf",ans);
    return 0;
}

^_^