一个c++程序，用于检查两条给定线段是否相交

#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 10th slides from following link for derivation of the formula
    // http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
    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;
}

问题可以简化成这样一个问题:从A到B和从C到D的两条直线相交吗?然后你可以问它四次(在直线和矩形的四条边之间)。

这是做这个的矢量数学。假设A到B的直线就是问题中的直线C到D的直线是其中一条矩形直线。我的表示法是Ax是A的x坐标Cy是c的y坐标“*”表示点积，例如A*B = Ax*Bx + Ay*By。

E = B-A = ( Bx-Ax, By-Ay )
F = D-C = ( Dx-Cx, Dy-Cy ) 
P = ( -Ey, Ex )
h = ( (A-C) * P ) / ( F * P )

h是键。如果h在0和1之间，两条线相交，否则不相交。如果F*P为零，当然不能进行计算，但在这种情况下，直线是平行的，因此只有在明显的情况下才相交。

交点是C + F*h。

更多的乐趣:

如果h恰好等于0或1，两条直线的端点相交。你可以认为这是一个“交集”，也可以认为不是。

具体来说，h是直线长度乘以多少才能恰好与另一条直线相交。

因此，如果h<0，这意味着矩形线在给定直线的“后面”(“方向”是“从A到B”)，如果h>1，矩形线在给定直线的“前面”。

推导:

A和C是指向直线起点的向量;E和F是由A和C端点组成的直线。

对于平面上任意两条不平行线，必须恰好有一对标量g和h，使得这个方程成立:

A + E*g = C + F*h

为什么?因为两条不平行线必须相交，这意味着你可以将这两条线按一定比例缩放并相互接触。

(起初，这看起来像一个有两个未知数的方程!但当你考虑到这是一个二维矢量方程时，它就不是，这意味着这是一对x和y的方程)

我们必须消去其中一个变量。一个简单的方法是使E项为零。要做到这一点，用一个向量对方程两边做点积这个向量与E点乘到0，我把上面的向量称为P，我做了E的明显变换。

你现在有:

A*P = C*P + F*P*h
(A-C)*P = (F*P)*h
( (A-C)*P ) / (F*P) = h

我已经尝试实现上述Jason所描述的算法;不幸的是，虽然在调试数学工作，我发现许多情况下，它不起作用。

例如，考虑点A(10,10) B(20,20) C(10,1) D(1,10) h=。5然而，通过检查可以清楚地看到，这些部分彼此一点也不接近。

将其绘制成图可以清楚地看出，0 < h < 1条件仅表明如果存在截距点，则截距点将位于CD上，而不告诉我们该点是否位于AB上。 为了确保有一个交叉点，你必须对变量g进行对称计算，拦截的要求是: 0 < g < 1 AND 0 < h < 1

基于@Gareth Rees的回答，Python版本:

import numpy as np

def np_perp( a ) :
    b = np.empty_like(a)
    b[0] = a[1]
    b[1] = -a[0]
    return b

def np_cross_product(a, b):
    return np.dot(a, np_perp(b))

def np_seg_intersect(a, b, considerCollinearOverlapAsIntersect = False):
    # https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect/565282#565282
    # http://www.codeproject.com/Tips/862988/Find-the-intersection-point-of-two-line-segments
    r = a[1] - a[0]
    s = b[1] - b[0]
    v = b[0] - a[0]
    num = np_cross_product(v, r)
    denom = np_cross_product(r, s)
    # If r x s = 0 and (q - p) x r = 0, then the two lines are collinear.
    if np.isclose(denom, 0) and np.isclose(num, 0):
        # 1. If either  0 <= (q - p) * r <= r * r or 0 <= (p - q) * s <= * s
        # then the two lines are overlapping,
        if(considerCollinearOverlapAsIntersect):
            vDotR = np.dot(v, r)
            aDotS = np.dot(-v, s)
            if (0 <= vDotR  and vDotR <= np.dot(r,r)) or (0 <= aDotS  and aDotS <= np.dot(s,s)):
                return True
        # 2. If neither 0 <= (q - p) * r = r * r nor 0 <= (p - q) * s <= s * s
        # then the two lines are collinear but disjoint.
        # No need to implement this expression, as it follows from the expression above.
        return None
    if np.isclose(denom, 0) and not np.isclose(num, 0):
        # Parallel and non intersecting
        return None
    u = num / denom
    t = np_cross_product(v, s) / denom
    if u >= 0 and u <= 1 and t >= 0 and t <= 1:
        res = b[0] + (s*u)
        return res
    # Otherwise, the two line segments are not parallel but do not intersect.
    return None

有一个很好的方法来解决这个问题就是用向量叉乘。定义二维向量叉乘v × w为vx wy−vy wx。

假设这两条线段从p到p + r，从q到q + s。那么第一行上的任意点都可以表示为p + t r(对于标量参数t)，第二行上的任意点可以表示为q + u s(对于标量参数u)。

如果t和u满足以下条件，两条直线相交:

P + t r = q + u s

两边叉乘s，得到

(p + r) × s = (q + u s) × s

由于s × s = 0，这意味着

T (r × s) = (q−p) × s

因此，求解t:

T = (q−p) × s / (r × s)

同样地，我们可以解出u:

(p + r) × r = (q + u s) × r U (s × r) = (p−q) × r U = (p−q) × r / (s × r)

为了减少计算步骤，可以方便地将其重写为以下形式(记住s × r =−r × s):

U = q−p × r / (r × s)

现在有四种情况:

If r × s = 0 and (q − p) × r = 0, then the two lines are collinear. In this case, express the endpoints of the second segment (q and q + s) in terms of the equation of the first line segment (p + t r): t0 = (q − p) · r / (r · r) t1 = (q + s − p) · r / (r · r) = t0 + s · r / (r · r) If the interval between t0 and t1 intersects the interval [0, 1] then the line segments are collinear and overlapping; otherwise they are collinear and disjoint. Note that if s and r point in opposite directions, then s · r < 0 and so the interval to be checked is [t1, t0] rather than [t0, t1]. If r × s = 0 and (q − p) × r ≠ 0, then the two lines are parallel and non-intersecting. If r × s ≠ 0 and 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1, the two line segments meet at the point p + t r = q + u s. Otherwise, the two line segments are not parallel but do not intersect.

来源:该方法是3D线相交算法的2维专门化，来自Ronald Goldman发表在Graphics Gems，第304页的文章“三条线在三维空间中的相交”。在三维空间中，通常的情况是直线是倾斜的(既不平行也不相交)，在这种情况下，该方法给出了两条直线最接近的点。

一个c++程序，用于检查两条给定线段是否相交

#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 10th slides from following link for derivation of the formula
    // http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
    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;
}