二维几何小结

struct Point{    double x , y;    Point(double x=0,double y=0):x(x),y(y) {}//构造函数，方便代码编写};typedef Point Vector ;//从程序实现上，Vector只是Point的别名//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);}//点-点=向量Vector operator - (Point A , Point 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);}const double eps = 1e-10;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;}向量(x,y)极角=atan2(y,x);  //单位弧度//两向量点积double Dot(Vector A , Vector B){    return A.x*B.x+A.y*B.y;}//向量长度double Length(Vector A){    return sqrt(Dot(A,A));}//两向量转角double Angle(Vector A , Vector B){    return acos(Dot(A, B) / Length(A) / Length(B));}//向量叉积double Cross(Vector A , Vector B){    return A.x*B.y - A.y*B.x;}//三角形有向面积两倍double Area2(Point A ,Point B ,Point C){    return Cross(B-A, C-A);}//向量绕起点旋转  rad是弧度Vector Rotate(Vector A, double rad){    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}//向量单位法向量Vector Normal(Vector A){    double L = Length(A);    return Vector(-A.y/L , A.x/L);}//二直线交点（参数式） 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;}//点到直线距离double DistanceToLine(Point P, Point A, Point B){    Vector v1 = B-A, v2 = P - A;    return fabs(Cross(v1,v2) / Length(v1));}//点到线段距离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 if(dcmp(Dot(v1,v3)) > 0) return Length(v3);    else return fabs(Cross(v1,v2)) / Length(v1);}//多边形面积double ConvexPolygonArea(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;}