考虑一个矩形ABCD，我们得到侧面AD和BC(分别为p和q)的中点的坐标以及它们的长度L(AD = BC = L)。现在给定参数，我们需要打印4个点A，B，C和D的坐标。
例子：
Input : p = (1, 0)
q = (1, 2)
L = 2
Output : (0, 0), (0, 2), (2, 2), (2, 0)
Explanation:
The printed points form a rectangle which
satisfy the input constraints.
Input : p = (1, 1)
q = (-1, -1)
L = 2*sqrt(2)
Output : (0, 2), (-2, 0), (0, -2), (2, 0)
从问题陈述中可以得出3种情况：
- 矩形是水平的，即AD和BC平行于X轴
- 矩形是垂直的，即AD和BC平行于Y轴
- 矩形与轴倾斜一定角度
前两种情况很简单，可以使用基本几何轻松解决。对于第三种情况，我们需要应用一些数学概念来找到要点。
为了清楚起见，请考虑上图。我们有p和q的坐标。因此，我们可以找到AD和BC的斜率(因为pq垂直于AD)。一旦有了AD的斜率，就可以找到通过AD的直线方程。现在我们可以应用距离公式来获得沿X和Y轴的位移。
If slope of AD = m, then
m = (p.x- q.x)/(q.y - p.y)
and displacement along X axis, dx =
L/(2*sqrt(1+m*m))
Similarly, dy = m*L/(2*sqrt(1+m*m))
现在，我们可以通过简单地添加和减去相应获得的位移来简单地找到4个角的坐标。
下面是实现。
C++
// C++ program to find corner points of
// a rectangle using given length and middle
// points.
#include
using namespace std;
// Structure to represent a co-ordinate point
struct Point
{
float x, y;
Point()
{
x = y = 0;
}
Point(float a, float b)
{
x = a, y = b;
}
};
// This function receives two points and length
// of the side of rectangle and prints the 4
// corner points of the rectangle
void printCorners(Point p, Point q, float l)
{
Point a, b, c, d;
// horizontal rectangle
if (p.x == q.x)
{
a.x = p.x - (l/2.0);
a.y = p.y;
d.x = p.x + (l/2.0);
d.y = p.y;
b.x = q.x - (l/2.0);
b.y = q.y;
c.x = q.x + (l/2.0);
c.y = q.y;
}
// vertical rectangle
else if (p.y == q.y)
{
a.y = p.y - (l/2.0);
a.x = p.x;
d.y = p.y + (l/2.0);
d.x = p.x;
b.y = q.y - (l/2.0);
b.x = q.x;
c.y = q.y + (l/2.0);
c.x = q.x;
}
// slanted rectangle
else
{
// calculate slope of the side
float m = (p.x-q.x)/float(q.y-p.y);
// calculate displacements along axes
float dx = (l /sqrt(1+(m*m))) *0.5 ;
float dy = m*dx;
a.x = p.x - dx;
a.y = p.y - dy;
d.x = p.x + dx;
d.y = p.y + dy;
b.x = q.x - dx;
b.y = q.y - dy;
c.x = q.x + dx;
c.y = q.y + dy;
}
cout << a.x << ", " << a.y << " n"
<< b.x << ", " << b.y << "n";
<< c.x << ", " << c.y << " n"
<< d.x << ", " << d.y << "nn";
}
// Driver code
int main()
{
Point p1(1, 0), q1(1, 2);
printCorners(p1, q1, 2);
Point p(1, 1), q(-1, -1);
printCorners(p, q, 2*sqrt(2));
return 0;
} Java
// Java program to find corner points of
// a rectangle using given length and middle
// points.
class GFG
{
// Structure to represent a co-ordinate point
static class Point
{
float x, y;
Point()
{
x = y = 0;
}
Point(float a, float b)
{
x = a;
y = b;
}
};
// This function receives two points and length
// of the side of rectangle and prints the 4
// corner points of the rectangle
static void printCorners(Point p, Point q, float l)
{
Point a = new Point(), b = new Point(),
c = new Point(), d = new Point();
// horizontal rectangle
if (p.x == q.x)
{
a.x = (float) (p.x - (l / 2.0));
a.y = p.y;
d.x = (float) (p.x + (l / 2.0));
d.y = p.y;
b.x = (float) (q.x - (l / 2.0));
b.y = q.y;
c.x = (float) (q.x + (l / 2.0));
c.y = q.y;
}
// vertical rectangle
else if (p.y == q.y)
{
a.y = (float) (p.y - (l / 2.0));
a.x = p.x;
d.y = (float) (p.y + (l / 2.0));
d.x = p.x;
b.y = (float) (q.y - (l / 2.0));
b.x = q.x;
c.y = (float) (q.y + (l / 2.0));
c.x = q.x;
}
// slanted rectangle
else
{
// calculate slope of the side
float m = (p.x - q.x) / (q.y - p.y);
// calculate displacements along axes
float dx = (float) ((l / Math.sqrt(1 + (m * m))) * 0.5);
float dy = m * dx;
a.x = p.x - dx;
a.y = p.y - dy;
d.x = p.x + dx;
d.y = p.y + dy;
b.x = q.x - dx;
b.y = q.y - dy;
c.x = q.x + dx;
c.y = q.y + dy;
}
System.out.print((int)a.x + ", " + (int)a.y + " \n"
+ (int)b.x + ", " + (int)b.y + "\n"
+ (int)c.x + ", " + (int)c.y + " \n"
+ (int)d.x + ", " + (int)d.y + "\n");
}
// Driver code
public static void main(String[] args)
{
Point p1 = new Point(1, 0), q1 = new Point(1, 2);
printCorners(p1, q1, 2);
Point p = new Point(1, 1), q = new Point(-1, -1);
printCorners(p, q, (float) (2 * Math.sqrt(2)));
}
}
// This code contributed by Rajput-JiC#
// C# program to find corner points of
// a rectangle using given length and middle
// points.
using System;
class GFG
{
// Structure to represent a co-ordinate point
public class Point
{
public float x, y;
public Point()
{
x = y = 0;
}
public Point(float a, float b)
{
x = a;
y = b;
}
};
// This function receives two points and length
// of the side of rectangle and prints the 4
// corner points of the rectangle
static void printCorners(Point p, Point q, float l)
{
Point a = new Point(), b = new Point(),
c = new Point(), d = new Point();
// horizontal rectangle
if (p.x == q.x)
{
a.x = (float) (p.x - (l / 2.0));
a.y = p.y;
d.x = (float) (p.x + (l / 2.0));
d.y = p.y;
b.x = (float) (q.x - (l / 2.0));
b.y = q.y;
c.x = (float) (q.x + (l / 2.0));
c.y = q.y;
}
// vertical rectangle
else if (p.y == q.y)
{
a.y = (float) (p.y - (l / 2.0));
a.x = p.x;
d.y = (float) (p.y + (l / 2.0));
d.x = p.x;
b.y = (float) (q.y - (l / 2.0));
b.x = q.x;
c.y = (float) (q.y + (l / 2.0));
c.x = q.x;
}
// slanted rectangle
else
{
// calculate slope of the side
float m = (p.x - q.x) / (q.y - p.y);
// calculate displacements along axes
float dx = (float) ((l / Math.Sqrt(1 + (m * m))) * 0.5);
float dy = m * dx;
a.x = p.x - dx;
a.y = p.y - dy;
d.x = p.x + dx;
d.y = p.y + dy;
b.x = q.x - dx;
b.y = q.y - dy;
c.x = q.x + dx;
c.y = q.y + dy;
}
Console.Write((int)a.x + ", " + (int)a.y + " \n"
+ (int)b.x + ", " + (int)b.y + "\n"
+ (int)c.x + ", " + (int)c.y + " \n"
+ (int)d.x + ", " + (int)d.y + "\n");
}
// Driver code
public static void Main(String[] args)
{
Point p1 = new Point(1, 0), q1 = new Point(1, 2);
printCorners(p1, q1, 2);
Point p = new Point(1, 1), q = new Point(-1, -1);
printCorners(p, q, (float) (2 * Math.Sqrt(2)));
}
}
// This code has been contributed by 29AjayKumar输出：
0, 0
0, 2
2, 2
2, 0
0, 2
-2, 0
0, -2
2, 0
参考：
堆栈溢出