从给定的三个坐标中找到所有可能的坐标,以制作非零区域的平行四边形。
我们称A,B,C为三个给定点。我们只能有以下三种可能的情况:
(1) AB and AC are sides, and BC a diagonal
(2) AB and BC are sides, and AC a diagonal
(3) BC and AC are sides, and AB a diagonal
因此,我们可以说,如果给出三个坐标,则只有3个坐标是可能的,从中我们可以生成平行四边形。
为了证明这三点都是不同的,让我们假设这是错误的。在不失一般性的前提下,假设在AD和BC情况下得到的分数相等。
考虑以下等式的两个方程式的系统:
Bx + Cx - Ax = Ax + Cx - Bx
By + Cy - Ay = Ay + Cy - By
It can be simplified as-
Ax = Bx
Ay = By
而且我们有一个矛盾,因为所有的点A,B,C都是不同的。
例子:
Input : A = (0 0)
B = (1 0)
C = (0 1)
Output : 1 -1
-1 1
1 1
Input : A = (-1 -1)
B = (0 1)
C = (1 1)
Output : -2 -1
0 -1
2 3
由于两边相等,AD = BC,AB = CD,我们可以将缺失点(D)的坐标计算为:
AD = BC
(Dx - Ax, Dy - Ay) = (Cx - Bx, Cy - By)
Dx = Ax + Cx - Bx
Dy = Ay + Cy - By
对角线是AD和BC,CD和AB的情况以相同的方式处理。参考: https : //math.stackexchange.com/questions/1322535/how-many-different-parallelograms-can-be-drawn-if-given-three-co-ordinates-in-3d
下面是上述方法的实现:
C++
// C++ program to all possible points
// of a parallelogram
#include
using namespace std;
// main method
int main()
{
int ax = 5, ay = 0; //coordinates of A
int bx = 1, by = 1; //coordinates of B
int cx = 2, cy = 5; //coordinates of C
cout << ax + bx - cx << ", "
<< ay + by - cy <
Java
// Java program to all possible
// points of a parallelogram
public class ParallelogramPoints{
// Driver code
public static void main(String[] s)
{
int ax = 5, ay = 0; //coordinates of A
int bx = 1, by = 1; //coordinates of B
int cx = 2, cy = 5; //coordinates of C
System.out.println(ax + bx - cx + ", "
+ (ay + by - cy));
System.out.println(ax + cx - bx + ", "
+ (ay + cy - by));
System.out.println(cx + bx - ax + ", "
+ (cy + by - ax));
}
}
// This code is contributed by Prerna Saini
Python3
# Python3 program to find all possible points
# of a parallelogram
ax = 5
ay = 0 #coordinates of A
bx = 1
by = 1 #coordinates of B
cx = 2
cy = 5 #coordinates of C
print(ax + bx - cx, ", ", ay + by - cy)
print(ax + cx - bx, ", ", ay + cy - by)
print(cx + bx - ax, ", ", cy + by - ax)
C#
// C# program to all possible
// points of a parallelogram
using System;
public class ParallelogramPoints
{
// Driver code
public static void Main()
{
//coordinates of A
int ax = 5, ay = 0;
//coordinates of B
int bx = 1, by = 1;
//coordinates of C
int cx = 2, cy = 5;
Console.WriteLine(ax + bx - cx + ", "
+ (ay + by - cy));
Console.WriteLine(ax + cx - bx + ", "
+ (ay + cy - by));
Console.WriteLine(cx + bx - ax + ", "
+ (cy + by - ax));
}
}
// This code is contributed by vt_m.
PHP
输出:
4, -4
6, 4
-2, 1
时间复杂度: O(1)