• 计算几何的简单模板


    const double eps = 1e-8;
    
    struct Point
    {
        double x,y;
        Point(double tx = 0,double ty = 0) : x(tx),y(ty){}
    };
    typedef Point Vtor;
    //向量的加减乘除
    Vtor operator + (Vtor A,Vtor B) { return Vtor(A.x + B.x,A.y + B.y); }
    Vtor operator - (Point A,Point B) { return Vtor(A.x - B.x,A.y - B.y); }
    Vtor operator * (Vtor A,double p) { return Vtor(A.x*p,A.y*p); }
    Vtor operator / (Vtor A,double p) { return Vtor(A.x/p,A.y/p); }
    bool operator < (Point A,Point B) { return A.x < B.x || (A.x == B.x && A.y < B.y);}
    int dcmp(double x){ if (fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }
    bool operator == (Point A,Point B) {return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0; }
    //向量的点积,长度,夹角
    double Dot(Vtor A,Vtor B) { return A.x*B.x + A.y*B.y; }
    double Length(Vtor A) { return sqrt(Dot(A,A)); }
    double Angle(Vtor A,Vtor B) { return acos(Dot(A,B)/Length(A)/Length(B)); }
    //叉积,三角形面积
    double Cross(Vtor A,Vtor B) { return A.x*B.y - A.y*B.x; }
    double Area2(Point A,Point B,Point C) { return Cross(B - A,C - A); }
    //向量的旋转,求向量的单位法线(即左转90度,然后长度归一)
    Vtor Rotate(Vtor A,double rad){ return Vtor(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad)); }
    Vtor Normal(Vtor A)
    {
        double L = Length(A);
        return Vtor(-A.y/L, A.x/L);
    }
    //直线的交点
    Point GetLineIntersection(Point P,Vtor v,Point Q,Vtor w)
    {
        Vtor u = P - Q;
        double t = Cross(w,u)/Cross(v,w);
        return P + v*t;
    }
    //点到直线的距离
    double DistanceToLine(Point P,Point A,Point B)
    {
        Vtor v1 = B - A;
        return fabs(Cross(P - A,v1))/Length(v1);
    }
    //点到线段的距离
    double DistanceToSegment(Point P,Point A,Point B)
    {
        if (A == B) return Length(P - A);
        Vtor v1 =  B - A , v2 = P - A, v3 = P - B;
        if (dcmp(Dot(v1,v2)) < 0) return Length(v2);
        else if (dcmp(Dot(v1,v3)) > 0) return Length(v3);
        else return fabs(Cross(v1,v2))/Length(v1);
    }
    //点到直线的映射
    Point GetLineProjection(Point P,Point A,Point B)
    {
        Vtor v = B - A;
        return A + v*Dot(v,P - A)/Dot(v,v);
    }
    
    //判断线段是否规范相交
    bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
    {
        double c1 = Cross(a2 - a1,b1 - a1), c2 = Cross(a2 - a1,b2 - a1),
               c3 = Cross(b2 - b1,a1 - b1), c4 = Cross(b2 - b1,a2 - b1);
        return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;
    }
    //判断点是否在一条线段上
    bool OnSegment(Point P,Point a1,Point a2)
    {
        return dcmp(Cross(a1 - P,a2 - P)) == 0 && dcmp(Dot(a1 - P,a2 - P)) < 0;
    }
    //多边形面积
    double PolgonArea(Point *p,int n)
    {
        double area = 0;
        for (int i = 1; i < n - 1; ++i)
        area += Cross(p[i] - p[0],p[i + 1] - p[0]);
        return area/2;
    }
    

      

    和圆有关的计算

    struct Line
    {
        Point p;
        Vtor v;
        Line(Point p,Vtor v) : p(p),v(v){}
        Point point(double t) { return p + v*t; }
    };
    struct Circle
    {
        Point c;
        double r;
        Circle(Point tc,double tr) : c(tc),r(tr){}
        Point point(double a)
        {
            return Point(c.x + cos(a)*r + c.y + sin(a)*r);
        }
    };
    //判断圆与直线是否相交以及求出交点
    int getLineCircleIntersection(Line L,Circle C,double t1,double t2,vector<Point> &sol)
    {
        //注意sol没有清空哦
        double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
        double e = a*a + c*c , f = 2*(a*b + c*d),  g = b*b + d*d;
        double delta = f*f - 4*e*g;
        if (dcmp(delta) < 0) return 0;
        else if (dcmp(delta) == 0)
        {
            t1 = t2 = -f/(2*e);
            sol.push_back(L.point(t1));
            return 1;
        }
        t1 = (-f - sqrt(delta))/(2*e); sol.push_back(L.point(t1));
        t2 = (-f + sqrt(delta))/(2*e); sol.push_back(L.point(t2));
        return 2;
    }
    //判断并求出两圆的交点
    double angle(Vtor v) { return atan2(v.y, v.x); }
    int getCircleIntersection(Circle C1,Circle C2,vector<Point> &sol)
    {
        double d = Length(C2.c - C1.c);
        // 圆心重合
        if (dcmp(d) == 0)
        {
            if (dcmp(C1.r - C2.r) == 0) return -1; // 两圆重合
            return 0; // 包含
        }
    
        // 圆心不重合
        if (dcmp(C1.r + C2.r - d) < 0) return 0; // 相离
        if (dcmp(fabs(C1.r - C2.r) - d) > 0) return 0; // 包含
    
        double a = angle(C2.c - C1.c);
        double da = acos(C1.r*C1.r + d*d - C2.r*C2.r) / (2*C1.r*d);
        Point p1 = C1.point(a - da), p2 = C1.point(a + da);
        sol.push_back(p1);
        if (p1 == p2) return 1;
        sol.push_back(p2);
        return 2;
    }
    //求点到圆的切线
    int getTangents(Point p,Circle C,Vtor *v)
    {
        Vtor u = C.c - p;
        double dis = Length(u);
        if (dis < C.r)  return 0;
        else if (dcmp(dis - C.r) == 0)
        {
            v[0] = Rotate(u,PI/2);
            return 1;
        }
        else
        {
            double ang = asin(C.r / dis);
            v[0] = Rotate(u, -ang);
            v[1] = Rotate(u, ang);
            return 2;
        }
    }
    //求两圆的切线
    int getCircleTangents(Circle A,Circle B,Point *a,Point *b)
    {
        int cnt = 0;
        if (A.r < B.r) { swap(A,B); swap(a, b) ; }
        //圆心距的平方
        double d2 = (A.c.x - B.c.x)*(A.c.x - B.c.x) + (A.c.y - B.c.y)*(A.c.y - B.c.y);
        double rdiff = A.r - B.r;
        double rsum = A.r + B.r;
        double base = angle(B.c - A.c);
        //重合有无限多条
        if (d2 == 0 && dcmp(A.r - B.r) == 0) return -1;
        //内切
        if (dcmp(d2 - rdiff*rdiff) == 0)
        {
            a[cnt] = A.point(base);
            b[cnt] = B.point(base); cnt++;
            return 1;
        }
        //有外公切线
        double ang = acos((A.r - B.r) / sqrt(d2));
        a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
        a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;
    
        //一条内切线
        if (dcmp(d2 - rsum*rsum) == 0)
        {
            a[cnt] = A.point(base); b[cnt] = B.point(PI + base); cnt++;
        }//两条内切线
        else if (dcmp(d2 - rsum*rsum) > 0)
        {
            double ang = acos((A.r + B.r) / sqrt(d2));
            a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
            a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;
        }
        return cnt;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/E-star/p/3187318.html
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