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  • 【转】计算几何模板

    转载注明出处 @AOQNRMGYXLMV 】

    #include <cstdio>
    #include <algorithm>
    #include <cmath>
    #include <vector>
    using namespace std;
    //lrj计算几何模板
    struct Point
    {
        double x, y;
        Point(double x=0, double y=0) :x(x),y(y) {}
    };
    typedef Point Vector;
    
    Point read_point(void)
    {
        double x, y;
        scanf("%lf%lf", &x, &y);
        return Point(x, y);
    }
    
    const double EPS = 1e-10;
    
    //向量+向量=向量 点+向量=点
    Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }
    
    //向量-向量=向量 点-点=向量
    Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); }
    
    //向量*数=向量
    Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); }
    
    //向量/数=向量
    Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); }
    
    bool operator < (const Point& a, const 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 == (const Point& a, const Point& b)
    { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
    
    /**********************基本运算**********************/
    
    //点积
    double Dot(Vector A, Vector B)
    { return A.x*B.x + A.y*B.y; }
    //向量的模
    double Length(Vector A)    { return sqrt(Dot(A, A)); }
    
    //向量的夹角,返回值为弧度
    double Angle(Vector A, Vector B)
    { return acos(Dot(A, B) / Length(A) / Length(B)); }
    
    //叉积
    double Cross(Vector A, Vector B)
    { return A.x*B.y - A.y*B.x; }
    
    //向量AB叉乘AC的有向面积
    double Area2(Point A, Point B, Point C)
    { return Cross(B-A, C-A); }
    
    //向量A旋转rad弧度
    Vector VRotate(Vector A, double rad)
    {
        return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
    }
    
    //将B点绕A点旋转rad弧度
    Point PRotate(Point A, Point B, double rad)
    {
        return A + VRotate(B-A, rad);
    }
    
    //求向量A向左旋转90°的单位法向量,调用前确保A不是零向量
    Vector Normal(Vector A)
    {
        double l = Length(A);
        return Vector(-A.y/l, A.x/l);
    }
    
    /**********************点和直线**********************/
    
    //求直线P + tv 和 Q + tw的交点,调用前要确保两条直线有唯一交点
    Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)
    {
        Vector u = P - Q;
        double t = Cross(w, u) / Cross(v, w);
        return P + v*t;
    }//在精度要求极高的情况下,可以自定义分数类
    
    //P点到直线AB的距离
    double DistanceToLine(Point P, Point A, Point B)
    {
        Vector v1 = B - A, v2 = P - A;
        return fabs(Cross(v1, v2)) / Length(v1);    //不加绝对值是有向距离
    }
    
    //点到线段的距离
    double DistanceToSegment(Point P, Point A, Point B)
    {
        if(A == B)    return Length(P - A);
        Vector 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);
    }
    
    //点在直线上的射影
    Point GetLineProjection(Point P, Point A, Point B)
    {
        Vector v = B - A;
        return A + v * (Dot(v, P - A) / Dot(v, v));
    }
    
    //线段“规范”相交判定
    bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)
    {
        double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);
        double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);
        return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
    }
    
    //判断点是否在线段上
    bool OnSegment(Point P, Point a1, Point a2)
    {
        Vector v1 = a1 - P, v2 = a2 - P;
        return dcmp(Cross(v1, v2)) == 0 && dcmp(Dot(v1, v2)) < 0;
    }
    
    //求多边形面积
    double PolygonArea(Point* P, int n)
    {
        double ans = 0.0;
        for(int i = 1; i < n - 1; ++i)
            ans += Cross(P[i]-P[0], P[i+1]-P[0]);
        return ans/2;
    }
    
    int main(void)
    {
        Vector a[2];
        sort(a, a + 2);
        return 0;
    }
    /**********************圆的相关计算**********************/
    
    const double PI = acos(-1.0);
    struct Line
    {//有向直线
        Point p;
        Vector v;
        double ang;
        Line()    { }
        Line(Point p, Vector v): p(p), v(v)    { ang = atan2(v.y, v.x); }
        Point point(double t)
        {
            return p + v*t;
        }
        bool operator < (const Line& L) const
        {
            return ang < L.ang;
        }
    };
    
    struct Circle
    {
        Point c;    //圆心
        double r;    //半径
        Circle(Point c, double r):c(c), r(r)    {}
        Point point(double a)
        {//求对应圆心角的点
            return Point(c.x + r*cos(a), c.y + r*sin(a));
        }
    };
    
    //两圆相交并返回交点个数 
    int getLineCircleIntersection(Line L, Circle C, double& t1, double& t2, vector<Point>& sol)
    {
        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 - C.r*C.r;
        double delta = f*f - 4*e*g;        //判别式
        if(dcmp(delta) < 0)    return 0;    //相离
        if(dcmp(delta) == 0)            //相切
        {
            t1 = t2 = -f / (2 * e);
            sol.push_back(L.point(t1));
            return 1;
        }
        //相交
        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;
    }
    
    //计算向量极角
    double angle(Vector v)    { return atan2(v.y, v.x); }
    
    int getCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol)
    {//圆与圆相交,并返回交点个数
        double d = Length(C1.c - C2.c);
        if(dcmp(d) == 0)
        {
            if(dcmp(C1.r - C2.r) == 0)    return -1;    //两圆重合
            return 0;                                //没有交点
        }
        if(dcmp(C1.r + C2.r - d) > 0)    return 0;
        if(dcmp(fabs(C1.r - C2.r) - d) > 0)    return 0;
    
        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;
    }
    
    //过定点作圆的切线并返回切线条数
    int getTangents(Point p, Circle C, Vector* v)
    {
        Vector u = C.c - p;
        double dist = Length(u);
        if(dist < C.r)    return 0;
        else if(dcmp(dist - C.r) == 0)
        {
            v[0] = VRotate(u, PI/2);
            return 1;
        }
        else
        {
            double ang = asin(C.r / dist);
            v[0] = VRotate(u, +ang);
            v[1] = VRotate(u, -ang);
            return 2;
        }
    }
    
    //求两个圆的公切线,并返回切线条数
    //注意,这里的Circle和上面的定义的Circle不一样
    int getTangents(Circle A, Circle B, Point* a, Point* b)
    {
        int cnt = 0;
        if(A.r < B.r)    { swap(A, B); swap(a, b); }
        double d2 = (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
        double rdiff = A.r - B.r;
        double rsum = A.r + B.r;
        if(d2 < rdiff*rdiff)    return 0;    //内含
    
        double base = atan2(B.y-A.y, B.x-A.x);
        if(dcmp(d2) == 0 && dcmp(A.r - B.r) == 0)    return -1; //重合
        if(dcmp(d2 - rdiff*rdiff) == 0)    //内切
        {
            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++;
        if(dcmp(rsum*rsum - d2) == 0)
        {//外切
            a[cnt] = b[cnt] = A.point(base); cnt++;
        }
        else if(dcmp(d2 - rsum*rsum) > 0)
        {
            ang = acos((A.r + B.r) / sqrt(d2));
            a[cnt] = A.point(base + ang); b[cnt] = B.point(PI + base + ang); cnt++;
            a[cnt] = A.point(base - ang); b[cnt] = B.point(PI + base - ang); cnt++;
        }
        return cnt;
    }
    
    //转角发判定点P是否在多边形内部
    int isPointInPolygon(Point P, Point* Poly, int n)
    {
        int wn;
        for(int i = 0; i < n; ++i)
        {
            if(OnSegment(P, Poly[i], Poly[(i+1)%n]))    return -1;    //在边界上
            int k = dcmp(Cross(Poly[(i+1)%n] - Poly[i], P - Poly[i]));
            int d1 = dcmp(Poly[i].y - P.y);
            int d2 = dcmp(Poly[(i+1)%n].y - P.y);
            if(k > 0 && d1 <= 0 && d2 > 0)    wn++;
            if(k < 0 && d2 <= 0 && d1 > 0)    wn--;
        }
        if(wn != 0)    return 1;    //内部
        return 0;                //外部
    }
    
    //计算凸包,输入点数组P,个数为n,输出点数组ch。函数返回凸包顶点数。
    //输入不能有重复点,函数执行后点的顺序会发生变化
    //如果不希望凸包的边上有输入点,把两个 <= 改成 <
    //在精度要求高时,可用dcmp比较
    int ConvexHull(Point* p, int n, Point* ch)
    {
        sort(p, p +n);
        int m = 0;
        for(int i = 0; i < n; ++i)
        {
            while(m > 1 && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0)    m--;
            ch[m++] = p[i];
        }
        int k = m;
        for(int i = n-2; i >= 0; --i)
        {
            while(m > k && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0)    m--;
            ch[m++] = p[i];
        }
        if(n > 1)    m--;
        return m;
    }
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  • 原文地址:https://www.cnblogs.com/zhyfzy/p/4298667.html
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