Problem G: 平面上的点和线——Point类、Line类 (VII)
Problem G: 平面上的点和线——Point类、Line类 (VII)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 4968 Solved: 3403
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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::showCounter()方法:按格式输出当前程序中Point对象的计数。
Point::showSum()方法:按格式输出程序运行至当前存在过的Point对象总数。
Line::showCounter()方法:按格式输出当前程序中Line对象的计数。
Line::showSum()方法:按格式输出程序运行至当前存在过的Line对象总数。
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
Current : 3 points.
In total : 3 points.
Current : 6 lines.
In total : 6 lines.
Current : 17 points.
In total : 17 points.
Current : 6 lines.
In total : 7 lines.
Current : 15 points.
In total : 17 points.
Current : 6 lines.
In total : 8 lines.
Current : 17 points.
In total : 21 points.
Current : 6 lines.
In total : 9 lines.
Current : 15 points.
In total : 21 points.
Current : 6 lines.
In total : 10 lines.
Current : 17 points.
In total : 25 points.
Current : 6 lines.
In total : 11 lines.
Current : 15 points.
In total : 25 points.
Current : 6 lines.
In total : 12 lines.
Current : 17 points.
In total : 29 points.
Current : 6 lines.
In total : 13 lines.
Current : 15 points.
In total : 29 points.
Current : 9 lines.
In total : 17 lines.
Current : 21 points.
In total : 37 points.
Current : 13 lines.
In total : 21 lines.
Current : 21 points.
In total : 45 points.
#include <iostream> using namespace std; class Point { friend class Line; private: double st,ed; static int now,sum; public: Point():st(0),ed(0){sum++;now++;} Point(double stt ,double edd ):st(stt),ed(edd){sum++;now++;} Point (const Point &p) {st = p.st;ed= p.ed; sum++;now++;} double x() const {return st;} double y() const {return ed;} ~Point (){now--;} static void showCounter(){cout<<"Current : "<<now<<" points."<<endl;} static void showSum(){cout<<"In total : "<<sum<<" points."<<endl;} }; int Point::sum=0; int Point::now=0; class Line { friend class Point; private : Point st_,ed_; public: static int s; static int n; Line():st_(0,0),ed_(0,0) {s++;n++;} Line(Point &sst, Point &eed):st_(sst),ed_(eed) {s++;n++;} Line(double x1,double y1,double x2,double y2):st_(x1,y1),ed_(x2,y2) {s++;n++;} ~Line() { n--;} Line(const Line &l):st_(l.st_),ed_(l.ed_) {s++;n++;} Line setLine(double x1,double y1,double x2,double y2) { st_.st=x1; st_.ed=y1; ed_.st=x2; ed_.ed=y2; return *this;} Line &setLine(const Point &p ,const Point &q) { st_=p; ed_=q; return *this; } Line &setLine(const Line &p) { st_=p.st_; ed_=p.ed_; return *this; } void readLine() { double x1,y1,x2,y2; char in; cin>>x1>>in>>y1>>x2>>in>>y2; st_.st=x1; st_.ed=y1; ed_.st=x2; ed_.ed=y2; } /*const Point &start()const {return st_;} const Point &end()const {return ed_;} void setStart(Point &s) {st_=s;} void setEnd(Point &e) {ed_=e;}*/ static void showCounter(){cout<<"Current : "<<n<<" lines."<<endl;} static void showSum(){cout<<"In total : "<<s<<" lines."<<endl;} }; int Line::s=0; int Line::n=0; int main() { int num, i; Point p(1, -2), q(2, -1), t; t.showCounter(); t.showSum(); std::cin>>num; Line line[num + 1]; for(i = 1; i <= num; i++) { Line *l1, l2; l1->showCounter(); l1->showSum(); l1 = new Line(p, q); line[i].readLine(); p.showCounter(); p.showSum(); delete l1; l2.showCounter(); l2.showSum(); q.showCounter(); q.showSum(); } Line l1(p, q), l2(p,t), l3(q,t), l4(l1); Line::showCounter(); Line::showSum(); Point::showCounter(); Point::showSum(); Line *l = new Line[num]; l4.showCounter(); l4.showSum(); delete[] l; t.showCounter(); t.showSum(); }