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  • Luogu-2600 [ZJOI2008]瞭望塔

    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
    ",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
    ",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;
    }
    
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  • 原文地址:https://www.cnblogs.com/nianheng/p/10013873.html
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