C++ 3 个有序点的方向（Orientation of 3 ordered points）

news2024/11/17 13:53:28

给定三个点 p1、p2 和 p3，任务是确定这三个点的方向。

平面中有序三重点的方向可以是

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，则 (p3, p2, p1) 的方向也共线。

如果 (p1, p2, p3) 的方向是顺时针，则 (p3, p2, p1) 的方向是逆时针，反之亦然。

例子：

输入： p1 = {0, 0}, p2 = {4, 4}, p3 = {1, 2}

输出： 逆时针

输入： p1 = {0, 0}, p2 = {4, 4}, p3 = {1, 1}

输出： 共线

如何计算方向？

这个想法就是利用斜率。

线段 (p1, p2) 的斜率：？ = (y2 - y1)/(x2 - x1)

线段 (p2, p3) 的斜率：？ = (y3 - y2)/(x3 - x2)

如果 ? > ?，则方向为顺时针（右转）

使用上述 ? 和 ? 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 ? < ? 时，上述表达式为负数，即逆时针

下面是上述想法的实现：

// A C++ program to find orientation of three points
#include <iostream>
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 collinear
// 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; // collinear

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";
cout << endl;

p1 = { 0, 0 }, p2 = { 4, 4 }, p3 = { 1, 1 };
o = orientation(p1, p2, p3);
if (o == 0)
cout << "Linear";
else if (o == 1)
cout << "Clockwise";
else
cout << "CounterClockwise";
cout << endl;

p1 = { 1, 2 }, p2 = { 4, 4 }, p3 = { 0, 0 };
o = orientation(p1, p2, p3);
if (o == 0)
cout << "Linear";
else if (o == 1)
cout << "Clockwise";
else
cout << "CounterClockwise";
return 0;
}

输出：

逆时针

线性

顺时针

时间复杂度： O(1)

辅助空间： O(1)

以下文章使用了方向的概念：

找到给定点集的简单闭合路径

c++ https://blog.csdn.net/hefeng_aspnet/article/details/141715783
java https://blog.csdn.net/hefeng_aspnet/article/details/141715877
python https://blog.csdn.net/hefeng_aspnet/article/details/141715899
C# https://blog.csdn.net/hefeng_aspnet/article/details/141715914
Javascript https://blog.csdn.net/hefeng_aspnet/article/details/141715931

如何检查两个给定的线段是否相交？

c++ https://blog.csdn.net/hefeng_aspnet/article/details/141713655
java https://blog.csdn.net/hefeng_aspnet/article/details/141713762
python https://blog.csdn.net/hefeng_aspnet/article/details/141714389
C# https://blog.csdn.net/hefeng_aspnet/article/details/141714420
Javascript https://blog.csdn.net/hefeng_aspnet/article/details/141714442

凸包 | 集合 1（贾维斯算法或包装）

c++ https://blog.csdn.net/hefeng_aspnet/article/details/141716082
java https://blog.csdn.net/hefeng_aspnet/article/details/141716363
python https://blog.csdn.net/hefeng_aspnet/article/details/141716444
C# https://blog.csdn.net/hefeng_aspnet/article/details/141716403
Javascript https://blog.csdn.net/hefeng_aspnet/article/details/141716421

凸包 | 集合 2（格雷厄姆扫描）

c++ https://blog.csdn.net/hefeng_aspnet/article/details/141716566
java https://blog.csdn.net/hefeng_aspnet/article/details/141717095
python https://blog.csdn.net/hefeng_aspnet/article/details/141717139
C# https://blog.csdn.net/hefeng_aspnet/article/details/141717214
Javascript https://blog.csdn.net/hefeng_aspnet/article/details/141717264

方法＃2：使用斜率

该方法通过计算由点形成的线段的斜率来检查平面中 3 个有序点的方向。如果斜率相等，则这些点共线。如果前两个点形成的线段的斜率小于后两个点形成的线段的斜率，则方向为逆时针，否则为顺时针。

算法
1. 计算 (p1,p2) 和 (p2,p3) 形成的线的斜率

2. 如果斜率相等，则点共线

3. 如果 (p1,p2) 的斜率 < (p2,p3) 的斜率，则点处于逆时针方向

4. 如果 (p1,p2) 的斜率 > (p2,p3) 的斜率，则点处于顺时针方向

#include <iostream>
using namespace std;

string orientation(int p1[], int p2[], int p3[])
{
// Calculate slopes
int slope1 = (p2[1] - p1[1]) * (p3[0] - p2[0]);
int slope2 = (p3[1] - p2[1]) * (p2[0] - p1[0]);
// Check orientation
if (slope1 == slope2) {
return "Collinear";
}
else if (slope1 < slope2) {
return "CounterClockWise";
}
else {
return "ClockWise";
}
}

int main()
{
// Example usage
int p1[] = { 0, 0 };
int p2[] = { 4, 4 };
int p3[] = { 1, 1 };
cout << orientation(p1, p2, p3) << endl;

int p4[] = { 0, 0 };
int p5[] = { 4, 4 };
int p6[] = { 1, 2 };
cout << orientation(p4, p5, p6) << endl;

return 0;
}
// This code is contributed by user_dtewbxkn77n