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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