第十二周上机项目4 点，圆的关系

【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）；

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;  }  

/** Copyright (c) 2015,烟台大学计算机学院* All right reserved.* 作者：曹莉萍* 文件：Demo.cpp* 完成时间：2015年05月31日* 版本号：v1.0*/#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;}

（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。

/** Copyright (c) 2015,烟台大学计算机学院* All right reserved.* 作者：曹莉萍* 文件：Demo.cpp* 完成时间：2015年05月31日* 版本号：v1.0*/#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。

方法1：用引用类型参数

/** Copyright (c) 2015,烟台大学计算机学院* All right reserved.* 作者：曹莉萍* 文件：Demo.cpp* 完成时间：2015年05月31日* 版本号：v1.0*/#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;}

方法2：结构体（定义两个包含两个点的结构体，用于返回值）

/** Copyright (c) 2015,烟台大学计算机学院* All right reserved.* 作者：曹莉萍* 文件：Demo.cpp* 完成时间：2015年05月31日* 版本号：v1.0*/#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;}

好复杂的项目4，千辛万苦弄完以后竟然有种字符搬家的感觉。

赶快吸收为自己的吧。

加油加油！