［几何］判断两个线段是否相交(多语言实现)

本文主要讲怎么判断两个线段是否相交

参考博客：
http://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/

http://www.cise.ufl.edu/~sitharam/COURSES/CG/kreveldintrolinesegment.pdf

http://geomalgorithms.com/a09-_intersect-3.html
http://hsfzxjy.github.io/the-simplest-way-to-find-out-if-two-segments-are-intersected/

How to check if two given line segments intersect?

Given two line segments (p1, q1) and (p2, q2), find if
the given line segments intersect with each other.

Before we discuss solution, let us define notion of orientation. Orientation
of an ordered triplet of points in the plane can be

–counterclockwise

–clockwise

–colinear

The following diagram shows different possible orientations of (a, b, c)



How is Orientation useful here?

Two segments (p1,q1) and (p2,q2) intersect if and only
if one of the following two conditions is verified

1. General Case:

– (p1, q1, p2)
and (p1, q1, q2)
have different orientations and

– (p2, q2, p1)
and (p2, q2, q1)
have different orientations.

Examples:



2. Special Case 

– (p1, q1, p2),
(p1, q1, q2),
(p2, q2, p1),
and (p2, q2, q1)
are all collinear and

– the x-projections of (p1, q1) and (p2, q2)
intersect

– the y-projections of (p1, q1) and (p2, q2)
intersect

Examples:



可以使用java.awt.geom.Line2D的朋友，请直接使用系统API: intersectsLine (http://docs.oracle.com/javase/7/docs/api/java/awt/geom/Line2D.html)

// intersectsLine
public boolean intersectsLine(double x1, double y1, double x2, double y2)
Tests if the line segment from (x1,y1) to (x2,y2) intersects this line segment.
// Parameters:
x1 - the X coordinate of the start point of the specified line segment
y1 - the Y coordinate of the start point of the specified line segment
x2 - the X coordinate of the end point of the specified line segment
y2 - the Y coordinate of the end point of the specified line segment
// Returns:
if this line segment and the specified line segment intersect each other; false otherwise.
// Since:
1.2

不能使用系统API的小伙伴，我们只能自己造轮子了。

C++语言实现：

// A C++ program to check if two given line segments intersect
#include <iostream>
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 http://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 p2 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;
}

输出结果：

No
Yes
No

Java语言实现：

// Given three colinear points p, q, r, the function checks if point q lies on line segment 'pr'
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
*
* @param p
* @param q
* @param r
* @return 0 --> p, q and r are colinear, 1 --> 顺时针方向, 2 --> 逆时钟方向
*/
int orientation(Point p, Point q, Point r) {
// See http://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.
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 p2 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
}