如何检查两个给定的线段是否相交？

给定两条线段(p1, q1)和(p2, q2)，找出给定的线段是否相交。
在我们讨论解决方案之前，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

方向如何在这里有用？
当且仅当验证以下两个条件之一时，两个线段(p1,q1)和(p2,q2)相交
1.一般情况：
– ( p1 , q1 , p2 ) 和 ( p1 , q1 , q2 ) 有不同的方向和
– ( p2 , q2 , p1 ) 和 ( p2 , q2 , q1 ) 有不同的方向。
例子：

2.特例
– ( p1 , q1 , p2 ), ( p1 , q1 , q2 ), ( p2 , q2 , p1 ) 和 ( p2 , q2 , q1 ) 都共线并且
– ( p1 , q1 ) 和 ( p2 , q2 ) 的 x 投影相交
– ( p1 , q1 ) 和 ( p2 , q2 ) 的 y 投影相交
例子：

以下是基于上述想法的实现。

C++

// A C++ program to check if two given line segments intersect
#include 
using namespace std;
 
struct Point
{
    int x;
    int y;
};
 
// Given three colinear 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 colinear
// 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;  // colinear
 
    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 colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;
 
    // p1, q1 and q2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;
 
    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;
 
     // p2, q2 and q1 are colinear 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 colinear 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 colinear
// 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; // colinear
 
    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 colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;
 
    // p1, q1 and q2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;
 
    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;
 
    // p2, q2 and q1 are colinear 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 colinear 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 : Colinear 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:
         
        # Colinear 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 colinear and p2 lies on segment p1q1
    if ((o1 == 0) and onSegment(p1, p2, q1)):
        return True
 
    # p1 , q1 and q2 are colinear and q2 lies on segment p1q1
    if ((o2 == 0) and onSegment(p1, q2, q1)):
        return True
 
    # p2 , q2 and p1 are colinear and p1 lies on segment p2q2
    if ((o3 == 0) and onSegment(p2, p1, q2)):
        return True
 
    # p2 , q2 and q1 are colinear 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#

// 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 colinear 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 colinear
// 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; // colinear
 
    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 colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;
 
    // p1, q1 and q2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;
 
    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;
 
    // p2, q2 and q1 are colinear 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

// Javascript program to check if two given line segments intersect
 
class Point
{
    constructor(x, y)
    {
        this.x = x;
            this.y = y;
    }
}
 
// Given three colinear 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 colinear
// 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; // colinear
   
    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 colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;
   
    // p1, q1 and q2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;
   
    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;
   
    // p2, q2 and q1 are colinear 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

输出：

No
Yes
No

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