如何判断两个线段是否相交呢？

2022-09-11 乐帮网

数学 javascript c# java c++

原文：https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
给定两条线段 (p1, q1) 和 (p2, q2)，求给定的线段是否彼此相交。
在讨论解决方案之前，让我们定义方向的概念。平面中有序三元组的方向可以是
--逆时针
--顺时针
--共线

下图显示了 (a, b, c) 的不同可能方向

定向在这里有什么用处？
当且仅当以下两个条件之一得到验证时，两个线段 (p1,q1) 和 (p2,q2) 相交，

一、一般情况：
– (p1, q1, p2) 和 (p1, q1, q2) 具有不同的方向和
– (p2, q2, p1) 和 (p2, q2, q1) 具有不同的方向。

定向在这里有什么用处？
当且仅当以下两个条件之一得到验证时，两个线段 (p1,q1) 和 (p2,q2) 相交

2. 特殊情况
– (p1, q1, p2), (p1, q1, q2), (p2, q2, p1) 和 (p2, q2, q1) 都是共线的并且
– (p1, q1) 和 (p2, q2) 的 x 投影相交
– (p1, q1) 和 (p2, q2) 的 y 投影相交

以下是基于上述思想的实现

C++

// A C++ program to check if two given line segments intersect
#include <iostream>
using namespace std;

struct Point
{
	int x;
	int y;
};

// Given three collinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
bool onSegment(Point p, Point q, Point r)
{
	if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
		q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
	return true;

	return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are collinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
	// See https://www.geeksforgeeks.org/orientation-3-ordered-points/
	// for details of below formula.
	int val = (q.y - p.y) * (r.x - q.x) -
			(q.x - p.x) * (r.y - q.y);

	if (val == 0) return 0; // collinear

	return (val > 0)? 1: 2; // clock or counterclock wise
}

// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
	// Find the four orientations needed for general and
	// special cases
	int o1 = orientation(p1, q1, p2);
	int o2 = orientation(p1, q1, q2);
	int o3 = orientation(p2, q2, p1);
	int o4 = orientation(p2, q2, q1);

	// General case
	if (o1 != o2 && o3 != o4)
		return true;

	// Special Cases
	// p1, q1 and p2 are collinear and p2 lies on segment p1q1
	if (o1 == 0 && onSegment(p1, p2, q1)) return true;

	// p1, q1 and q2 are collinear and q2 lies on segment p1q1
	if (o2 == 0 && onSegment(p1, q2, q1)) return true;

	// p2, q2 and p1 are collinear and p1 lies on segment p2q2
	if (o3 == 0 && onSegment(p2, p1, q2)) return true;

	// p2, q2 and q1 are collinear and q1 lies on segment p2q2
	if (o4 == 0 && onSegment(p2, q1, q2)) return true;

	return false; // Doesn't fall in any of the above cases
}

// Driver program to test above functions
int main()
{
	struct Point p1 = {1, 1}, q1 = {10, 1};
	struct Point p2 = {1, 2}, q2 = {10, 2};

	doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

	p1 = {10, 0}, q1 = {0, 10};
	p2 = {0, 0}, q2 = {10, 10};
	doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

	p1 = {-5, -5}, q1 = {0, 0};
	p2 = {1, 1}, q2 = {10, 10};
	doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

	return 0;
}

Java

// Java program to check if two given line segments intersect
class GFG
{

static class Point
{
	int x;
	int y;

		public Point(int x, int y)
		{
			this.x = x;
			this.y = y;
		}
	
};

// Given three collinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
static boolean onSegment(Point p, Point q, Point r)
{
	if (q.x <= Math.max(p.x, r.x) && q.x >= Math.min(p.x, r.x) &&
		q.y <= Math.max(p.y, r.y) && q.y >= Math.min(p.y, r.y))
	return true;

	return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are collinear
// 1 --> Clockwise
// 2 --> Counterclockwise
static int orientation(Point p, Point q, Point r)
{
	// See https://www.geeksforgeeks.org/orientation-3-ordered-points/
	// for details of below formula.
	int val = (q.y - p.y) * (r.x - q.x) -
			(q.x - p.x) * (r.y - q.y);

	if (val == 0) return 0; // collinear

	return (val > 0)? 1: 2; // clock or counterclock wise
}

// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
static boolean doIntersect(Point p1, Point q1, Point p2, Point q2)
{
	// Find the four orientations needed for general and
	// special cases
	int o1 = orientation(p1, q1, p2);
	int o2 = orientation(p1, q1, q2);
	int o3 = orientation(p2, q2, p1);
	int o4 = orientation(p2, q2, q1);

	// General case
	if (o1 != o2 && o3 != o4)
		return true;

	// Special Cases
	// p1, q1 and p2 are collinear and p2 lies on segment p1q1
	if (o1 == 0 && onSegment(p1, p2, q1)) return true;

	// p1, q1 and q2 are collinear and q2 lies on segment p1q1
	if (o2 == 0 && onSegment(p1, q2, q1)) return true;

	// p2, q2 and p1 are collinear and p1 lies on segment p2q2
	if (o3 == 0 && onSegment(p2, p1, q2)) return true;

	// p2, q2 and q1 are collinear and q1 lies on segment p2q2
	if (o4 == 0 && onSegment(p2, q1, q2)) return true;

	return false; // Doesn't fall in any of the above cases
}

// Driver code
public static void main(String[] args)
{
	Point p1 = new Point(1, 1);
	Point q1 = new Point(10, 1);
	Point p2 = new Point(1, 2);
	Point q2 = new Point(10, 2);

	if(doIntersect(p1, q1, p2, q2))
		System.out.println("Yes");
	else
		System.out.println("No");

	p1 = new Point(10, 1); q1 = new Point(0, 10);
	p2 = new Point(0, 0); q2 = new Point(10, 10);
	if(doIntersect(p1, q1, p2, q2))
			System.out.println("Yes");
	else
		System.out.println("No");

	p1 = new Point(-5, -5); q1 = new Point(0, 0);
	p2 = new Point(1, 1); q2 = new Point(10, 10);;
	if(doIntersect(p1, q1, p2, q2))
		System.out.println("Yes");
	else
		System.out.println("No");
}
}

// This code is contributed by Princi Singh

Python3

# A Python3 program to find if 2 given line segments intersect or not

class Point:
	def __init__(self, x, y):
		self.x = x
		self.y = y

# Given three collinear points p, q, r, the function checks if
# point q lies on line segment 'pr'
def onSegment(p, q, r):
	if ( (q.x <= max(p.x, r.x)) and (q.x >= min(p.x, r.x)) and
		(q.y <= max(p.y, r.y)) and (q.y >= min(p.y, r.y))):
		return True
	return False

def orientation(p, q, r):
	# to find the orientation of an ordered triplet (p,q,r)
	# function returns the following values:
	# 0 : Collinear points
	# 1 : Clockwise points
	# 2 : Counterclockwise
	
	# See https://www.geeksforgeeks.org/orientation-3-ordered-points/amp/
	# for details of below formula.
	
	val = (float(q.y - p.y) * (r.x - q.x)) - (float(q.x - p.x) * (r.y - q.y))
	if (val > 0):
		
		# Clockwise orientation
		return 1
	elif (val < 0):
		
		# Counterclockwise orientation
		return 2
	else:
		
		# Collinear orientation
		return 0

# The main function that returns true if
# the line segment 'p1q1' and 'p2q2' intersect.
def doIntersect(p1,q1,p2,q2):
	
	# Find the 4 orientations required for
	# the general and special cases
	o1 = orientation(p1, q1, p2)
	o2 = orientation(p1, q1, q2)
	o3 = orientation(p2, q2, p1)
	o4 = orientation(p2, q2, q1)

	# General case
	if ((o1 != o2) and (o3 != o4)):
		return True

	# Special Cases

	# p1 , q1 and p2 are collinear and p2 lies on segment p1q1
	if ((o1 == 0) and onSegment(p1, p2, q1)):
		return True

	# p1 , q1 and q2 are collinear and q2 lies on segment p1q1
	if ((o2 == 0) and onSegment(p1, q2, q1)):
		return True

	# p2 , q2 and p1 are collinear and p1 lies on segment p2q2
	if ((o3 == 0) and onSegment(p2, p1, q2)):
		return True

	# p2 , q2 and q1 are collinear and q1 lies on segment p2q2
	if ((o4 == 0) and onSegment(p2, q1, q2)):
		return True

	# If none of the cases
	return False

# Driver program to test above functions:
p1 = Point(1, 1)
q1 = Point(10, 1)
p2 = Point(1, 2)
q2 = Point(10, 2)

if doIntersect(p1, q1, p2, q2):
	print("Yes")
else:
	print("No")

p1 = Point(10, 0)
q1 = Point(0, 10)
p2 = Point(0, 0)
q2 = Point(10,10)

if doIntersect(p1, q1, p2, q2):
	print("Yes")
else:
	print("No")

p1 = Point(-5,-5)
q1 = Point(0, 0)
p2 = Point(1, 1)
q2 = Point(10, 10)

if doIntersect(p1, q1, p2, q2):
	print("Yes")
else:
	print("No")
	
# This code is contributed by Ansh Riyal

// C# program to check if two given line segments intersect
using System;
using System.Collections.Generic;

class GFG
{

public class Point
{
	public int x;
	public int y;

	public Point(int x, int y)
	{
		this.x = x;
		this.y = y;
	}
	
};

// Given three collinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
static Boolean onSegment(Point p, Point q, Point r)
{
	if (q.x <= Math.Max(p.x, r.x) && q.x >= Math.Min(p.x, r.x) &&
		q.y <= Math.Max(p.y, r.y) && q.y >= Math.Min(p.y, r.y))
	return true;

	return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are collinear
// 1 --> Clockwise
// 2 --> Counterclockwise
static int orientation(Point p, Point q, Point r)
{
	// See https://www.geeksforgeeks.org/orientation-3-ordered-points/
	// for details of below formula.
	int val = (q.y - p.y) * (r.x - q.x) -
			(q.x - p.x) * (r.y - q.y);

	if (val == 0) return 0; // collinear

	return (val > 0)? 1: 2; // clock or counterclock wise
}

// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
static Boolean doIntersect(Point p1, Point q1, Point p2, Point q2)
{
	// Find the four orientations needed for general and
	// special cases
	int o1 = orientation(p1, q1, p2);
	int o2 = orientation(p1, q1, q2);
	int o3 = orientation(p2, q2, p1);
	int o4 = orientation(p2, q2, q1);

	// General case
	if (o1 != o2 && o3 != o4)
		return true;

	// Special Cases
	// p1, q1 and p2 are collinear and p2 lies on segment p1q1
	if (o1 == 0 && onSegment(p1, p2, q1)) return true;

	// p1, q1 and q2 are collinear and q2 lies on segment p1q1
	if (o2 == 0 && onSegment(p1, q2, q1)) return true;

	// p2, q2 and p1 are collinear and p1 lies on segment p2q2
	if (o3 == 0 && onSegment(p2, p1, q2)) return true;

	// p2, q2 and q1 are collinear and q1 lies on segment p2q2
	if (o4 == 0 && onSegment(p2, q1, q2)) return true;

	return false; // Doesn't fall in any of the above cases
}

// Driver code
public static void Main(String[] args)
{
	Point p1 = new Point(1, 1);
	Point q1 = new Point(10, 1);
	Point p2 = new Point(1, 2);
	Point q2 = new Point(10, 2);

	if(doIntersect(p1, q1, p2, q2))
		Console.WriteLine("Yes");
	else
		Console.WriteLine("No");

	p1 = new Point(10, 1); q1 = new Point(0, 10);
	p2 = new Point(0, 0); q2 = new Point(10, 10);
	if(doIntersect(p1, q1, p2, q2))
			Console.WriteLine("Yes");
	else
		Console.WriteLine("No");

	p1 = new Point(-5, -5); q1 = new Point(0, 0);
	p2 = new Point(1, 1); q2 = new Point(10, 10);;
	if(doIntersect(p1, q1, p2, q2))
		Console.WriteLine("Yes");
	else
		Console.WriteLine("No");
}
}

/* This code contributed by PrinciRaj1992 */

JavaScript

<script>
// Javascript program to check if two given line segments intersect

class Point
{
	constructor(x, y)
	{
		this.x = x;
			this.y = y;
	}
}

// Given three collinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
function onSegment(p, q, r)
{
	if (q.x <= Math.max(p.x, r.x) && q.x >= Math.min(p.x, r.x) &&
		q.y <= Math.max(p.y, r.y) && q.y >= Math.min(p.y, r.y))
	return true;
	
	return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are collinear
// 1 --> Clockwise
// 2 --> Counterclockwise
function orientation(p, q, r)
{

	// See https://www.geeksforgeeks.org/orientation-3-ordered-points/
	// for details of below formula.
	let val = (q.y - p.y) * (r.x - q.x) -
			(q.x - p.x) * (r.y - q.y);
	
	if (val == 0) return 0; // collinear
	
	return (val > 0)? 1: 2; // clock or counterclock wise
}

// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
function doIntersect(p1, q1, p2, q2)
{

	// Find the four orientations needed for general and
	// special cases
	let o1 = orientation(p1, q1, p2);
	let o2 = orientation(p1, q1, q2);
	let o3 = orientation(p2, q2, p1);
	let o4 = orientation(p2, q2, q1);
	
	// General case
	if (o1 != o2 && o3 != o4)
		return true;
	
	// Special Cases
	// p1, q1 and p2 are collinear and p2 lies on segment p1q1
	if (o1 == 0 && onSegment(p1, p2, q1)) return true;
	
	// p1, q1 and q2 are collinear and q2 lies on segment p1q1
	if (o2 == 0 && onSegment(p1, q2, q1)) return true;
	
	// p2, q2 and p1 are collinear and p1 lies on segment p2q2
	if (o3 == 0 && onSegment(p2, p1, q2)) return true;
	
	// p2, q2 and q1 are collinear and q1 lies on segment p2q2
	if (o4 == 0 && onSegment(p2, q1, q2)) return true;
	
	return false; // Doesn't fall in any of the above cases
}

// Driver code
let p1 = new Point(1, 1);
let q1 = new Point(10, 1);
let p2 = new Point(1, 2);
let q2 = new Point(10, 2);

if(doIntersect(p1, q1, p2, q2))
	document.write("Yes<br>");
else
	document.write("No<br>");

p1 = new Point(10, 1); q1 = new Point(0, 10);
p2 = new Point(0, 0); q2 = new Point(10, 10);
if(doIntersect(p1, q1, p2, q2))
	document.write("Yes<br>");
else
	document.write("No<br>");

p1 = new Point(-5, -5); q1 = new Point(0, 0);
p2 = new Point(1, 1); q2 = new Point(10, 10);;
if(doIntersect(p1, q1, p2, q2))
	document.write("Yes<br>");
else
	document.write("No<br>");

// This code is contributed by avanitrachhadiya2155
</script>

Output:

No
Yes
No

Sources:
http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
Introduction to Algorithms 3rd Edition by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest

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