3个有序点的方向 - 芒果文档

平面中有序三元组点的方向可以是

逆时针
顺时针
共线

下图显示了 (a,b,c) 的不同可能方向

如果 (p1, p2, p3) 的方向共线，那么 (p3, p2, p1) 的方向也共线。
如果 (p1, p2, p3) 的方向是顺时针，那么 (p3, p2, p1) 的方向是逆时针，反之亦然。

示例：给定三个点 p1、p2 和 p3，找出 (p1, p2, p3) 的方向。

Input:   p1 = {0, 0}, p2 = {4, 4}, p3 = {1, 2}
Output:  CounterClockWise

Input:   p1 = {0, 0}, p2 = {4, 4}, p3 = {1, 1}
Output:  Colinear

如何计算方向？

这个想法是使用斜率。线段(p1, p2)的斜率：σ = (y2 – y1)/(x2 – x1) 线段(p2, p3)的斜率：τ = (y3 – y2)/(x3 – x2) 如果σ > τ ，方向是顺时针（右转）使用上述 σ 和 τ 的值，我们可以得出结论，方向取决于以下表达式的符号： (y2 – y1)*(x3 – x2) – (y3 – y2)*( x2 – x1) 当 σ < τ 时，上式为负，即逆时针

下面是上述想法的实现。

C++

// A C++ program to find orientation of three points
#include 
using namespace std;
  
struct Point
{
    int x, y;
};
  
// To find orientation of ordered triplet (p1, p2, p3).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p1, Point p2, Point p3)
{
    // See 10th slides from following link for derivation
    // of the formula
    int val = (p2.y - p1.y) * (p3.x - p2.x) -
              (p2.x - p1.x) * (p3.y - p2.y);
  
    if (val == 0) return 0;  // colinear
  
    return (val > 0)? 1: 2; // clock or counterclock wise
}
  
// Driver program to test above functions
int main()
{
    Point p1 = {0, 0}, p2 = {4, 4}, p3 = {1, 2};
    int o = orientation(p1, p2, p3);
    if (o==0)         cout << "Linear";
    else if (o == 1)  cout << "Clockwise";
    else              cout << "CounterClockwise";
    return 0;
}

Java

// JAVA Code to find Orientation of 3
// ordered points
class Point
{
    int x, y;
    Point(int x,int y){
        this.x=x;
        this.y=y;
    }
}
  
class GFG {
      
    // To find orientation of ordered triplet 
    // (p1, p2, p3). The function returns 
    // following values 
    // 0 --> p, q and r are colinear
    // 1 --> Clockwise
    // 2 --> Counterclockwise
    public static int orientation(Point p1, Point p2,
                                         Point p3)
    {
        // See 10th slides from following link 
        // for derivation of the formula
        int val = (p2.y - p1.y) * (p3.x - p2.x) -
                  (p2.x - p1.x) * (p3.y - p2.y);
       
        if (val == 0) return 0;  // colinear
       
        // clock or counterclock wise
        return (val > 0)? 1: 2; 
    }
      
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
            Point p1 = new Point(0, 0);
            Point p2 = new Point(4, 4);
            Point p3 = new Point(1, 2);
              
            int o = orientation(p1, p2, p3);
              
            if (o==0)     
            System.out.print("Linear");
            else if (o == 1)  
            System.out.print("Clockwise");
            else              
            System.out.print("CounterClockwise");
          
    }
}
  
//This code is contributed by Arnav Kr. Mandal.

Python3

# A Python3 program to find orientation of 3 points
class Point:
      
    # to store the x and y coordinates of a point
    def __init__(self, x, y):
        self.x = x
        self.y = y
  
def orientation(p1, p2, p3):
      
    # to find the orientation of 
    # an ordered triplet (p1,p2,p3)
    # function returns the following values:
    # 0 : Colinear points
    # 1 : Clockwise points
    # 2 : Counterclockwise
    val = (float(p2.y - p1.y) * (p3.x - p2.x)) - \
           (float(p2.x - p1.x) * (p3.y - p2.y))
    if (val > 0):
          
        # Clockwise orientation
        return 1
    elif (val < 0):
          
        # Counterclockwise orientation
        return 2
    else:
          
        # Colinear orientation
        return 0
  
# Driver code
p1 = Point(0, 0)
p2 = Point(4, 4)
p3 = Point(1, 2)
  
o = orientation(p1, p2, p3)
  
if (o == 0):
    print("Linear")
elif (o == 1):
    print("Clockwise")
else:
    print("CounterClockwise")
      
# This code is contributed by Ansh Riyal

C#

// C# Code to find Orientation of 3
// ordered points
using System;
public class Point
{
    public int x, y;
    public Point(int x,int y)
    {
        this.x = x;
        this.y = y;
    }
}
  
class GFG 
{
      
    // To find orientation of ordered triplet 
    // (p1, p2, p3). The function returns 
    // following values 
    // 0 --> p, q and r are colinear
    // 1 --> Clockwise
    // 2 --> Counterclockwise
    public static int orientation(Point p1, Point p2,
                                        Point p3)
    {
        // See 10th slides from following link 
        // for derivation of the formula
        int val = (p2.y - p1.y) * (p3.x - p2.x) -
                (p2.x - p1.x) * (p3.y - p2.y);
      
        if (val == 0) return 0; // colinear
      
        // clock or counterclock wise
        return (val > 0)? 1: 2; 
    }
      
    /* Driver program to test above function */
    public static void Main(String[] args) 
    {
            Point p1 = new Point(0, 0);
            Point p2 = new Point(4, 4);
            Point p3 = new Point(1, 2);
              
            int o = orientation(p1, p2, p3);
              
            if (o == 0)     
                Console.WriteLine("Linear");
            else if (o == 1) 
                Console.WriteLine("Clockwise");
            else            
                Console.WriteLine("CounterClockwise");
          
    }
}
  
/* This code contributed by PrinciRaj1992 */

输出：

CounterClockwise

方向的概念在以下文章中使用：

找到一组给定点的简单闭合路径

如何检查两个给定的线段是否相交？

凸包 |设置 1（Jarvis 的算法或包装）

凸包 |第 2 组（格雷厄姆扫描）

资料来源： http : //www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf

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