我的代码只从文件中读入Points
,然后按自然顺序对它们进行排序(首先比较y坐标,然后比较x坐标),然后按点的斜率对第二点进行排序。 / p>
对于第一次排序,我重载了<
运算符并调用了sort();
对于第二次排序,我创建了一个由第二点初始化的函数对象。
我重写了Point
的复制构造函数,以找出任何不必要的复制,并发现我复制了第二次Point
,但我无法理解为什么。任何人都能给我一个线索吗?
输出:
C:\Users\lenovo\Desktop>test.exe < input10.txt
During initialization : 0
Input reading has complete!
Sort by natural order : (28000,1000) (28000,5000) (28000,13500) (23000,16000) (1000,
18000) (13000,21000) (2000,22000) (3000,26000) (3500,28000) (4000,30000)
During soring : 0
Sort by slope : copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
copying28000,13500
(28000,13500) (4000,30000) (3500,28000) (23000,16000) (13000,21000) (3000,26000) (20
00,22000) (1000,18000) (28000,1000) (28000,5000)
During soring : 12
代码:
#include <iostream>
#include <iterator>
#include <vector>
#include <map>
#include <list>
#include <string>
#include <algorithm>
using namespace std;
int times;
class Point
{
private:
int x,y;
public:
Point() : x(0),y(0){}
Point(int x,int y):x(x),y(y){}
Point(const Point& p) : x(p.x),y(p.y) { cout << "copying" <<x<<","<<y<<endl;times++;}
double slopeTo(const Point& that) const
{
if (x == that.x && y == that.y) return - numeric_limits<double>::infinity();
if (x == that.x) return numeric_limits<double>::infinity();
if (y == that.y) return 0;
return (that.y - y) / (double)(that.x - x);
}
bool operator< (const Point& that)const
{
if (y < that.y) return true;
if (y == that.y && x < that.x) return true;6
return false;
}
friend ostream& operator<< (ostream&, const Point& p);
};
class cmpBySlope
{
private:
Point origin;
public:
cmpBySlope(Point& a) : origin(a){}
bool operator() (const Point* left,const Point* right)const
{
return origin.slopeTo(*left) < origin.slopeTo(*right);
}
};
ostream& operator<< (ostream& out, const Point& p)
{
cout << "(" << p.x << "," << p.y << ")" ;
return out;
}
int N;
vector<Point*> v;
void create()
{
cin >> N;
for (int i = 0 ; i < N; i++)
{
int x,y;
cin >> x >> y;
Point* p = new Point(x,y);
}
cout << "During initialization : " << times << endl;
cout << "Input reading has complete!" << endl;
}
bool cmp(const Point* p,const Point* q)
{
return (*p)<(*q);
}
int main(void)
{
create();
int before = times;
cout << "Sort by natural order : ";
sort(v.begin(),v.end(),cmp);
for (Point* p : v)
cout << *p << " ";
cout << endl;
cout << "During soring : " << (times - before) << endl;
cout << "Sort by slope : ";
before = times;
sort(v.begin(),v.end(),cmpBySlope(*v[2]));
for (Point* p : v)
{
cout << *p << " ";
}
cout << endl;
cout << "During soring : " << (times - before) << endl;
}
答案 0 :(得分:0)
如评论中所示,Point副本来自cmpBySlope对象,该对象包含一个完整的Point实例。多次复制cmpBySlope的事实是sort函数的实现细节。
看一下这个answer,看看sort函数可能会使用什么算法。在这两种情况下,都使用递归,因为Compare对象是按值传递的,所以必须要创建副本。