因此,我为自己制作了一个点式印刷课程,该课程应该让用户输入2元组。即x和y,然后按^ order将它们打印回用户,其中order表示p1 =(x,y)
#include <iostream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;
class Point2D {
public:
Point2D();
Point2D(double a, double b);
double getx();
double gety();
void setx(double a);
void sety(double b);
virtual void print();
virtual void print(int a);
double angle();
private:
double x;
double y;
};
bool operator<( Point2D a , Point2D b );
int main() {
double my_x=-999;
double my_y=-999;
string my_color;
double my_weight;
vector<Point2D*> points;
cout << "Welcome to Point Printer! Please insert the x-and y-coordinates for your points and I will print them in sorted order! Just one rule, the point (0,0) is reserved as the terminating point, so when you are done enter (0,0).\n";
while(true)
{
cout << "x = ";
cin>>my_x;
cout << "y = ";
cin>>my_y;
if((my_x == 0)&&(my_y==0))
{
break;
}
points.push_back(new Point2D(my_x, my_y));
}
sort(points.begin(), points.end());
cout << "\n\n";
cout << "Your points are\n\n";
for(int i=0;i<points.size();i++)
{
cout<<i+1<<": ";
(*points[i]).print(); cout<<endl; // this is the printing gadget
}
for(int i=0; i<points.size(); i++)
{
delete points[i];
}
cout << endl << endl;
return 0;
}
double Point2D::angle()
{
double Angle = atan2(y,x);
if(Angle < 0)
{
return Angle + 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679;
}
return Angle;
}
bool operator< (Point2D a, Point2D b)
{
if (a.getx()*a.getx()+a.gety()*a.gety() < b.getx()*b.getx()+b.gety()*b.gety())
{
return true;
}
else if (a.getx()*a.getx()+a.gety()*a.gety() > b.getx()*b.getx()+b.gety()*b.gety())
{
return false;
}
if (a.getx()*a.getx()+a.gety()*a.gety() ==b.getx()*b.getx()+b.gety()*b.gety())
{
if (a.angle() < b.angle())
{
return true;
}
else if (a.angle() > b.angle())
{
return false;
}
}
return true;
}
Point2D::Point2D() { x = 0; y = 0; return;}
Point2D::Point2D(double a, double b) { x = a; y = b; return;}
double Point2D::getx() { return x;}
double Point2D::gety() { return y;}
void Point2D::setx(double a) { x = a; return; }
void Point2D::sety(double b) { y = b; return; }
void Point2D::print() {
cout<<"("<<x<<","<<y<<")";
return;
}
void Point2D::print(int a) {
print(); cout<<endl;
}
我遇到的问题是以下其中一种:
分类
角度()
运算符
完全不同...
特别要注意以下几点:
x = 1
y = 2
x = 2
y = 3
x = 1.1
y = 2.2
x = -10
y = 10
x = -5
y = -3
x = -5
y = 3
x = 5
y = -3
x = 5
y = 3
x = 0
y = 0
排序不正确。
任何帮助将非常感激。谢谢。
最佳答案
问题(或其中之一)是比较函数中的最终声明。
return true;
看一下这个块:
if (a.getx()*a.getx()+a.gety()*a.gety() ==b.getx()*b.getx()+b.gety()*b.gety())
{
if (a.angle() < b.angle())
{
return true;
}
else if (a.angle() > b.angle())
{
return false;
}
}
首先,如果到此为止,我们已经确定
(x*x + y*y)
和a
的b
计算是相等的。现在,我们假设角度也相等。怎么了?第一次测试失败,因为a.angle()
不小于b.angle()
。然后第二次测试失败,因为a.angle()
不大于b.angle()
。然后,您返回true。换句话说,您是说a
小于b
是正确的,即使按所有权利,也应将它们视为相等,因此应返回false。只需对return a.angle() < b.angle();
进行角度测试,而不是对角度进行多次测试,就可以解决问题。通过一些其他简化,您的函数应如下所示:bool operator<(Point2d a, Point2d b)
{
double A = a.getx()*a.getx()+a.gety()*a.gety();
double B = b.getx()*b.getx()+b.gety()*b.gety();
if (A < B) return true;
if (A > B) return false;
return a.angle() < b.angle();
}
关于c++ - 顺序:点排序分析,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/18201428/