C++第13周(春)项目3 - 与圆心相连的直线

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【项目3 - 与圆心相连的直线】
　　在项目1中定义的Point(点)类和Circle(圆)类基础上，设计一种方案，输出给定一点p与圆心相连成的直线与圆的两个交点。
　　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;
}

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