【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;
}
[参考解答]
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。 [参考解答]
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
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）
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;
}
