给定两个线段(p1，q1)和(p2，q2)，找出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。平面中点的有序三元组的方向可以是
–逆时针
-顺时针
–colinear
下图显示了( a ， b ， c )的不同可能方向
定向在这里有什么用?
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 )具有不同的方向，并且
–( p2 ， q2 ， p1 )和( p2 ， q2 ， q1 )具有不同的方向。
例子： 
2.特例
–( p1 ， q1 ， p2 )，( p1 ， q1 ， q2 )，( p2 ， q2 ， p1 )和( p2 ， q2 ， q1 )都是共线的，
–( p1 ， q1 )和( p2 ， q2 )的x投影相交
–( p1 ， q1 )和( p2 ， q2 )的y投影相交
例子： 
以下是基于以上思想的实现。
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 SinghPython3
# 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 RiyalC#
// 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 */输出：
No
Yes
No资料来源：
http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
算法入门第三版，作者：Clifford Stein，Thomas H. Cormen，Charles E. Leiserson，Ronald L. Rivest