【项目4 - 点、圆的关系】
(1)先建立一个Point(点)类,包含数据成员x,y(坐标点);
(2)以Point为基类,派生出一个Circle(圆)类,增加数据成员(半径),基类的成员表示圆心;
(3)编写上述两类中的构造、析构函数及必要运算符重载函数(本项目主要是输入输出);
(4)定义友元函数int locate,判断点p与圆的位置关系(返回值<0圆内,==0圆上,>0 圆外);
(5)在圆类上重载关系运算符(6种),使之能够按圆的面积比较两个圆的大小。自编main函数完成测试。
(1)先建立一个Point(点)类,包含数据成员x,y(坐标点);
(2)以Point为基类,派生出一个Circle(圆)类,增加数据成员(半径),基类的成员表示圆心;
(3)编写上述两类中的构造、析构函数及必要运算符重载函数(本项目主要是输入输出);
(4)定义友元函数int locate,判断点p与圆的位置关系(返回值<0圆内,==0圆上,>0 圆外);
int main( ) { Circle c1(3,2,4),c2(4,5,5); //c2应该大于c1 Point p1(1,1),p2(3,-2),p3(7,3); //分别位于c1内、上、外 cout<<"圆c1: "<<c1; cout<<"点p1: "<<p1; cout<<"点p1在圆c1之"<<((locate(p1, c1)>0)?"外":((locate(p1, c1)<0)?"内":"上"))<<endl; cout<<"点p2: "<<p2; cout<<"点p2在圆c1之"<<((locate(p2, c1)>0)?"外":((locate(p2, c1)<0)?"内":"上"))<<endl; cout<<"点p3: "<<p3; cout<<"点p3在圆c1之"<<((locate(p3, c1)>0)?"外":((locate(p3, c1)<0)?"内":"上"))<<endl; return 0; }
[参考解答]
#include <iostream> #include<Cmath> using namespace std; class Point { public: Point(double a=0,double b=0):x(a),y(b) {} //构造函数 double distance(const Point &p) const; //求距离 friend ostream & operator<<(ostream &,const Point &);//重载运算符“<<” protected: //受保护成员 double x,y; }; double Point::distance(const Point &p) const //求距离 { double dx = x-p.x; double dy = y-p.y; return sqrt(dx*dx+dy*dy); } ostream & operator<<(ostream &output,const Point &p) { output<<"["<<p.x<<","<<p.y<<"]"<<endl; return output; } class Circle:public Point //circle是Point类的公用派生类 { public: Circle(double a=0,double b=0,double r=0) :Point(a,b),radius(r) { }; //构造函数 friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<” friend int locate(const Point &p, const Circle &c); //判断点p在圆上、圆内或圆外,返回值:<0圆内,==0圆上,>0 圆外 protected: double radius; }; //重载运算符“<<”,使之按规定的形式输出圆的信息 ostream &operator<<(ostream &output,const Circle &c) { output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius<<endl; return output; } //判断点p在圆内、圆c内或圆c外 int locate(const Point &p, const Circle &c) { const Point cp(c.x,c.y); //圆心 double d = cp.distance(p); if (abs(d - c.radius) < 1e-7) return 0; //相等 else if (d < c.radius) return -1; //圆内 else return 1; //圆外 } int main( ) { Circle c1(3,2,4); Point p1(1,1),p2(3,-2),p3(7,3); //分别位于c1内、上、外 cout<<"圆c1: "<<c1; cout<<"点p1: "<<p1; cout<<"点p1在圆c1之"<<((locate(p1, c1)>0)?"外":((locate(p1, c1)<0)?"内":"上"))<<endl; cout<<"点p2: "<<p2; cout<<"点p2在圆c1之"<<((locate(p2, c1)>0)?"外":((locate(p2, c1)<0)?"内":"上"))<<endl; cout<<"点p3: "<<p3; cout<<"点p3在圆c1之"<<((locate(p3, c1)>0)?"外":((locate(p3, c1)<0)?"内":"上"))<<endl; return 0; }
[参考解答]
#include <iostream> #include<Cmath> using namespace std; class Point { public: Point(double a=0,double b=0):x(a),y(b) {} //构造函数 protected: //受保护成员 double x,y; }; class Circle:public Point //circle是Point类的公用派生类 { public: Circle(double a=0,double b=0,double r=0): Point(a,b),radius(r) { }//构造函数 double area ( ) const; //计算圆面积 friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<” //重载关系运算符运算符,使之能够按圆的面积比较两个圆的大小; bool operator>(const Circle &); bool operator<(const Circle &); bool operator>=(const Circle &); bool operator<=(const Circle &); bool operator==(const Circle &); bool operator!=(const Circle &); protected: double radius; }; //计算圆面积 double Circle::area( ) const { return 3.14159*radius*radius; } //重载运算符“<<”,使之按规定的形式输出圆的信息 ostream &operator<<(ostream &output,const Circle &c) { output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius; return output; } //重载关系运算符(种)运算符,使之能够按圆的面积比较两个圆的大小; bool Circle::operator>(const Circle &c) { return (this->radius - c.radius) > 1e-7; } bool Circle::operator<(const Circle &c) { return (c.radius - this->radius) > 1e-7; } bool Circle::operator>=(const Circle &c) { return !(*this < c); } bool Circle::operator<=(const Circle &c) { return !(*this > c); } bool Circle::operator==(const Circle &c) { return abs(this->radius - c.radius) < 1e-7; } bool Circle::operator!=(const Circle &c) { return abs(this->radius - c.radius) > 1e-7; } int main( ) { Circle c1(3,2,4),c2(4,5,5); //c2应该大于c1 cout<<"圆c1( "<<c1<<" )的面积是 "<<c1.area()<<endl; cout<<"圆c2( "<<c2<<" )的面积是 "<<c2.area()<<endl; cout<<"圆c1 "; if(c1>c2) cout<<"大于, "; if(c1<c2) cout<<"小于, "; if(c1>=c2) cout<<"大于等于, "; if(c1<=c2) cout<<"小于等于, "; if(c1==c2) cout<<"等于, "; if(c1!=c2) cout<<"不等于, "; cout<<"圆c2"<<endl; return 0; }
(6)与圆心相连的直线:给定一点p,其与圆心相连成的直线,会和圆有两个交点,如图。在上面定义的Point(点)类和Circle(圆)类基础上,设计一种方案,输出这两点的坐标。
提示:
[参考解答]
方案1:用引用类型参数获得结果
#include <iostream> #include<Cmath> using namespace std; class Circle; //由于在Point中声明友元函数crossover_point中参数中用了Circle,需要提前声明 class Point { public: Point(double a=0,double b=0):x(a),y(b) {} //构造函数 friend ostream & operator<<(ostream &,const Point &);//重载运算符“<<” friend void crossover_point(Point &p,Circle &c, Point &p1,Point &p2 ) ; //求交点的友元函数 protected: //受保护成员 double x,y; }; ostream & operator<<(ostream &output,const Point &p) { output<<"["<<p.x<<","<<p.y<<"]"; return output; } class Circle:public Point //circle是Point类的公用派生类 { public: Circle(double a=0,double b=0,double r=0):Point(a,b),radius(r) { } //构造函数 friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<” friend void crossover_point(Point &p,Circle &c, Point &p1,Point &p2 ) ; //求交点的友元函数 protected: double radius; }; //重载运算符“<<”,使之按规定的形式输出圆的信息 ostream &operator<<(ostream &output,const Circle &c) { output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius; return output; } //给定一点p,求出该点与圆c的圆心相连成的直线与圆的两个交点p1和p2 //关键问题是求得的交点如何返回 //方案1:利用引用类型的形式参数,注意,下面的p1和p2将“带回”求得的结果 //crossover_point函数已经声明为Point和Circle类的友元函数,类中私有成员可以直接访问 void crossover_point(Point &p, Circle &c, Point &p1,Point &p2 ) { p1.x = (c.x + sqrt(c.radius*c.radius/(1+((c.y-p.y)/(c.x-p.x))*((c.y-p.y)/(c.x-p.x))))); p2.x = (c.x - sqrt(c.radius*c.radius/(1+((c.y-p.y)/(c.x-p.x))*((c.y-p.y)/(c.x-p.x))))); p1.y = (p.y + (p1.x -p.x)*(c.y-p.y)/(c.x-p.x)); p2.y = (p.y + (p2.x -p.x)*(c.y-p.y)/(c.x-p.x)); } int main( ) { Circle c1(3,2,4); Point p1(1,1),p2,p3; crossover_point(p1,c1, p2, p3); cout<<"点p1: "<<p1<<endl; cout<<"与圆c1: "<<c1<<endl; cout<<"的圆心相连,与圆交于两点,分别是:"<<endl; cout<<"交点1: "<<p2<<endl; cout<<"交点2: "<<p3<<endl; return 0; }
方案2:定义一个包含两个点的结构体,专门用于返回值(定义成类进行封装可能更好)
#include <iostream> #include<Cmath> using namespace std; class Circle; //由于在Point中声明友元函数crossover_point中参数中用了Circle,需要提前声明 struct DoublePoint; //也先声明,Point中声明友元函数crossover_point中要用到 class Point { public: Point(double a=0,double b=0):x(a),y(b) {} //构造函数 friend ostream & operator<<(ostream &,const Point &);//重载运算符“<<” friend DoublePoint crossover_point(Point &p,Circle &c) ; //求交点的友元函数 protected: //受保护成员 double x,y; }; ostream & operator<<(ostream &output,const Point &p) { output<<"["<<p.x<<","<<p.y<<"]"; return output; } class Circle:public Point //circle是Point类的公用派生类 { public: Circle(double a=0,double b=0,double r=0):Point(a,b),radius(r) { } //构造函数 friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<” friend DoublePoint crossover_point(Point &p,Circle &c) ; //求交点的友元函数 protected: double radius; }; //重载运算符“<<”,使之按规定的形式输出圆的信息 ostream &operator<<(ostream &output,const Circle &c) { output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius; return output; } struct DoublePoint //专门用于返回值的结构体类型 { Point p1; Point p2; }; //给定一点p,求出该点与圆c的圆心相连成的直线与圆的两个交点 //方案2:结果返回到DoublePoint类型的结构体中 //crossover_point函数已经声明为Point和Circle类的友元函数,类中私有成员可以直接访问 DoublePoint crossover_point(Point &p, Circle &c) { DoublePoint pp; pp.p1.x = (c.x + sqrt(c.radius*c.radius/(1+((c.y-p.y)/(c.x-p.x))*((c.y-p.y)/(c.x-p.x))))); pp.p2.x = (c.x - sqrt(c.radius*c.radius/(1+((c.y-p.y)/(c.x-p.x))*((c.y-p.y)/(c.x-p.x))))); pp.p1.y = (p.y + (pp.p1.x -p.x)*(c.y-p.y)/(c.x-p.x)); pp.p2.y = (p.y + (pp.p2.x -p.x)*(c.y-p.y)/(c.x-p.x)); return pp; } int main( ) { Circle c1(3,2,4); Point p1(1,1); DoublePoint pp; pp = crossover_point(p1,c1); cout<<"点p1: "<<p1<<endl; cout<<"与圆c1: "<<c1<<endl; cout<<"的圆心相连,与圆交于两点,分别是:"<<endl; cout<<"交点1: "<<pp.p1<<endl; cout<<"交点2: "<<pp.p2<<endl; return 0; }