c++作业2-C（1329）平面上的点和线——Point类、Line类 (III)

Problem C: 平面上的点和线——Point类、Line类 (III)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 11684  Solved: 5974
[Submit][Status]

Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

根据“append.cc”，完成Point类和Line类的构造方法和show()方法，输出各Line对象和Point对象的构造和析构次序。

接口描述：

Point::show()方法：按格式输出Point对象。

Line::show()方法：按格式输出Line对象。

 

Input

输入的第一行为N，表示后面有N行测试样例。每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

 

Output

输出为多行，每行为一条线段，起点坐标在前终点坐标在后，每个点的X坐标在前，Y坐标在后，Y坐标前面多输出一个空格，用括号包裹起来。输出格式见sample。

C语言的输入输出被禁用。

 

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Point : (1, -2) is created. Point : (2, -1) is created. Point : (0, 0) is created. Point : (0, 0) ========================= Point : (0, 0) is created. Point : (1, 1) is created. Line : (0, 0) to (1, 1) is created. Line : (0, 0) to (1, 1) Line : (0, 0) to (1, 1) is erased. Point : (1, 1) is erased. Point : (0, 0) is erased. ========================= Point : (1, 1) is created. Point : (2, 3) is created. Line : (1, 1) to (2, 3) is created. Line : (1, 1) to (2, 3) Line : (1, 1) to (2, 3) is erased. Point : (2, 3) is erased. Point : (1, 1) is erased. ========================= Point : (2, 3) is created. Point : (4, 5) is created. Line : (2, 3) to (4, 5) is created. Line : (2, 3) to (4, 5) Line : (2, 3) to (4, 5) is erased. Point : (4, 5) is erased. Point : (2, 3) is erased. ========================= Point : (0, 1) is created. Point : (1, 0) is created. Line : (0, 1) to (1, 0) is created. Line : (0, 1) to (1, 0) Line : (0, 1) to (1, 0) is erased. Point : (1, 0) is erased. Point : (0, 1) is erased. ========================= Point : (1, -2) is copied. Point : (2, -1) is copied. Line : (1, -2) to (2, -1) is created. Point : (1, -2) is copied. Point : (0, 0) is copied. Line : (1, -2) to (0, 0) is created. Point : (2, -1) is copied. Point : (0, 0) is copied. Line : (2, -1) to (0, 0) is created. Point : (0, 0) is copied. Point : (2, -1) is copied. Line : (0, 0) to (2, -1) is created. Line : (1, -2) to (2, -1) Line : (1, -2) to (0, 0) Line : (2, -1) to (0, 0) Line : (0, 0) to (2, -1) Line : (0, 0) to (2, -1) is erased. Point : (2, -1) is erased. Point : (0, 0) is erased. Line : (2, -1) to (0, 0) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Line : (1, -2) to (0, 0) is erased. Point : (0, 0) is erased. Point : (1, -2) is erased. Line : (1, -2) to (2, -1) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased.

HINT

 

Append Code

append.cc,

int main()
{
    char c;
    int num, i;
    double x1, x2, y1, y2;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    for(i = 1; i <= num; i++)
    {
        std::cout<<"=========================\n";
        std::cin>>x1>>c>>y1>>x2>>c>>y2;
        Line line(x1, y1, x2, y2);
        line.show();
    }
    std::cout<<"=========================\n";
    Line l1(p, q), l2(p, t), l3(q, t), l4(t, q);
    l1.show();
    l2.show();
    l3.show();
    l4.show();
}

答案

#include<iostream>
using namespace std;
class Point
{
private:
    double x,y;
public:
    Point()
    {
        x=0;
        y=0;
        cout<<"Point : ("<<x<<", "<<y<<") is created."<<endl;
    }
    Point(double a,double b)
    {
        x=a;
        y=b;
        cout<<"Point : ("<<x<<", "<<y<<") is created."<<endl;
    }
    Point(const Point& p)
    {
        x=p.x;y=p.y;
        cout<<"Point : ("<<x<<", "<<y<<") is copied."<< endl;
    }
    ~Point()
    {
        cout<<"Point : ("<<x<<", "<<y<<") is erased."<<endl;
    }
    void show()
    {
        cout<<"Point : ("<<x<<", "<<y<<")"<< endl;
    }
    double getx()
    {
        return x;
    }
    double gety()
    {
        return y;
    }

};
class Line
{
private:
    Point p,q;
public:
    Line(Point& a,Point& b):p(a),q(b)
    {
        cout<<"Line : ("<<p.getx()<<", "<<p.gety()<<") to ("<<q.getx()<<", "<<q.gety()<<")"<<" is created."<< endl;
    }
    ~Line()
    {
        cout<<"Line : ("<<p.getx()<<", "<<p.gety()<<") to ("<<q.getx()<<", "<<q.gety()<<")"<<" is erased."<< endl;
    }
    void show()
    {
        cout<<"Line : ("<<p.getx()<<", "<<p.gety()<<") to ("<<q.getx()<<", "<<q.gety()<<")"<< endl;
    }
    Line(double a,double b,double c,double d):p(a,b),q(c,d)
    {
        cout<<"Line : ("<<p.getx()<<", "<<p.gety()<<") to ("<<q.getx()<<", "<<q.gety()<<")"<<" is created."<< endl;
    }
};
int main()
{
    char c;
    int num, i;
    double x1, x2, y1, y2;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    for(i = 1; i <= num; i++)
    {
        std::cout<<"=========================\n";
        std::cin>>x1>>c>>y1>>x2>>c>>y2;
        Line line(x1, y1, x2, y2);
        line.show();
    }
    std::cout<<"=========================\n";
    Line l1(p, q), l2(p, t), l3(q, t), l4(t, q);
    l1.show();
    l2.show();
    l3.show();
    l4.show();
}

疑惑：

1.为什么用Line(Point& a,Point& b)而不是Line(Point a,Point b)

2.为什么Line(Point a,Point b)会调用Point的拷贝构造四次且前两次顺序先b后a