• UVALive 3890 Most Distant Point from the Sea


    https://vjudge.net/problem/UVALive-3890

    题意:凸n边形小岛,求岛上距海的最远的一点距海的最短距离(凸包内切圆半径)

    二分+半平面交判断是否存在这样的点

    #include<cmath>
    #include<cstdio>
    #include<algorithm>
    
    #define N 101
    
    using namespace std;
    
    const double eps=1e-6;
    
    struct Point 
    {
        double x,y;
        
        Point (double x=0,double y=0) : x(x),y(y) { }
    };
    
    typedef Point Vector;
    
    Point P[N],Poly[N];
    Vector v[N],v2[N];
    
    Vector operator - (Vector A,Vector B) {  return Vector(A.x-B.x,A.y-B.y); }
    Vector operator + (Vector A,Vector B) {  return Vector(A.x+B.x,A.y+B.y); }
    Vector operator * (Vector A,double b) {  return Vector(A.x*b,A.y*b); }
    
    struct Line
    {
        Point P;
        Vector v;
        double ang;
        
        Line() {}
        Line(Point P,Vector v) : P(P),v(v) { ang=atan2(v.y,v.x); }
        
        bool operator < (Line L) const 
        {
            return ang<L.ang;
        }
    };
    
    Line L[N];
    
    double Cross(Vector A,Vector B)
    {
        return A.x*B.y-A.y*B.x;
    }
    
    double Dot(Vector A,Vector B)
    {
        return A.x*B.x+A.y*B.y;
    }
    
    double Length(Vector A)
    {
        return sqrt(Dot(A,A));
    }
    
    Vector Normal(Vector A)
    {
        double L=Length(A);
        return Vector(-A.y/L,A.x/L);
    }
    
    bool OnLeft(Line L,Point p)
    {
        return Cross(L.v,p-L.P)>0;
    }
    
    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;
    }
    
    bool HalfplaneIntersection(Line *L,int n)
    {
        sort(L,L+n);
        int first,last;
        Point *p=new Point[n];
        Line *q=new Line[n];
        q[first=last=0]=L[0];
        for(int i=1;i<n;++i)
        {
            while(first<last && !OnLeft(L[i],p[last-1])) last--;
            while(first<last && !OnLeft(L[i],p[first])) first++;
            q[++last]=L[i];
            if(fabs(Cross(q[last].v,q[last-1].v))<eps)
            {
                last--;
                if(OnLeft(q[last],L[i].P)) q[last]=L[i];
            }
            if(first<last) p[last-1]=GetIntersection(q[last-1],q[last]);
        }
        while(first<last && !OnLeft(q[first],p[last-1])) last--;
        return last-first>1;
    }
    
    int main()
    {
        int n,m;
        int x,y;
        double l,r,mid;
        while(scanf("%d",&n)!=EOF)
        {
            if(!n) return 0;
            for(int i=0;i<n;++i) 
            {
                scanf("%d%d",&x,&y);
                P[i]=Point(x,y);
            }
            for(int i=0;i<n;++i)
            {
                v[i]=P[(i+1)%n]-P[i];
                v2[i]=Normal(v[i]);
            }
            l=0; r=20000;
            while(r-l>eps)
            {
                mid=(l+r)/2;
                for(int i=0;i<n;++i) 
                L[i]=Line(P[i]+v2[i]*mid,v[i]);
                if(HalfplaneIntersection(L,n)) l=mid;
                else r=mid;
            }
            printf("%.6lf
    ",l);
        }
    }
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  • 原文地址:https://www.cnblogs.com/TheRoadToTheGold/p/8258798.html
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