如果矩形的每条边都是一条线段，并且用户绘制的部分也是一条线段，那么您只需检查用户绘制的线段是否与四条边线段相交。这应该是一个相当简单的练习，给定每个段的起点和终点。

这对我来说很有效。从这里拍的。

 // calculates intersection and checks for parallel lines.  
 // also checks that the intersection point is actually on  
 // the line segment p1-p2  
 Point findIntersection(Point p1,Point p2,  
   Point p3,Point p4) {  
   float xD1,yD1,xD2,yD2,xD3,yD3;  
   float dot,deg,len1,len2;  
   float segmentLen1,segmentLen2;  
   float ua,ub,div;  

   // calculate differences  
   xD1=p2.x-p1.x;  
   xD2=p4.x-p3.x;  
   yD1=p2.y-p1.y;  
   yD2=p4.y-p3.y;  
   xD3=p1.x-p3.x;  
   yD3=p1.y-p3.y;    

   // calculate the lengths of the two lines  
   len1=sqrt(xD1*xD1+yD1*yD1);  
   len2=sqrt(xD2*xD2+yD2*yD2);  

   // calculate angle between the two lines.  
   dot=(xD1*xD2+yD1*yD2); // dot product  
   deg=dot/(len1*len2);  

   // if abs(angle)==1 then the lines are parallell,  
   // so no intersection is possible  
   if(abs(deg)==1) return null;  

   // find intersection Pt between two lines  
   Point pt=new Point(0,0);  
   div=yD2*xD1-xD2*yD1;  
   ua=(xD2*yD3-yD2*xD3)/div;  
   ub=(xD1*yD3-yD1*xD3)/div;  
   pt.x=p1.x+ua*xD1;  
   pt.y=p1.y+ua*yD1;  

   // calculate the combined length of the two segments  
   // between Pt-p1 and Pt-p2  
   xD1=pt.x-p1.x;  
   xD2=pt.x-p2.x;  
   yD1=pt.y-p1.y;  
   yD2=pt.y-p2.y;  
   segmentLen1=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2);  

   // calculate the combined length of the two segments  
   // between Pt-p3 and Pt-p4  
   xD1=pt.x-p3.x;  
   xD2=pt.x-p4.x;  
   yD1=pt.y-p3.y;  
   yD2=pt.y-p4.y;  
   segmentLen2=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2);  

   // if the lengths of both sets of segments are the same as  
   // the lenghts of the two lines the point is actually  
   // on the line segment.  

   // if the point isn’t on the line, return null  
   if(abs(len1-segmentLen1)>0.01 || abs(len2-segmentLen2)>0.01)  
     return null;  

   // return the valid intersection  
   return pt;  
 }  

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

   void set(float x, float y){  
     this.x = x;  
     this.y = y;  
   }  
 }  

我尝试了很多方法，然后我决定自己写。就是这样:

bool IsBetween (float x, float b1, float b2)
{
   return ( ((x >= (b1 - 0.1f)) && 
        (x <= (b2 + 0.1f))) || 
        ((x >= (b2 - 0.1f)) &&
        (x <= (b1 + 0.1f))));
}

bool IsSegmentsColliding(   POINTFLOAT lineA,
                POINTFLOAT lineB,
                POINTFLOAT line2A,
                POINTFLOAT line2B)
{
    float deltaX1 = lineB.x - lineA.x;
    float deltaX2 = line2B.x - line2A.x;
    float deltaY1 = lineB.y - lineA.y;
    float deltaY2 = line2B.y - line2A.y;

    if (abs(deltaX1) < 0.01f && 
        abs(deltaX2) < 0.01f) // Both are vertical lines
        return false;
    if (abs((deltaY1 / deltaX1) -
        (deltaY2 / deltaX2)) < 0.001f) // Two parallel line
        return false;

    float xCol = (  (   (deltaX1 * deltaX2) * 
                        (line2A.y - lineA.y)) - 
                    (line2A.x * deltaY2 * deltaX1) + 
                    (lineA.x * deltaY1 * deltaX2)) / 
                 ((deltaY1 * deltaX2) - (deltaY2 * deltaX1));
    float yCol = 0;
    if (deltaX1 < 0.01f) // L1 is a vertical line
        yCol = ((xCol * deltaY2) + 
                (line2A.y * deltaX2) - 
                (line2A.x * deltaY2)) / deltaX2;
    else // L1 is acceptable
        yCol = ((xCol * deltaY1) +
                (lineA.y * deltaX1) -
                (lineA.x * deltaY1)) / deltaX1;

    bool isCol =    IsBetween(xCol, lineA.x, lineB.x) &&
            IsBetween(yCol, lineA.y, lineB.y) &&
            IsBetween(xCol, line2A.x, line2B.x) &&
            IsBetween(yCol, line2A.y, line2B.y);
    return isCol;
}

根据这两个公式:(由直线方程和其他公式简化而来)

我从《多视图几何》这本书里读到了这些算法

以下文本使用

'作为转置符号

*作为点积

当用作算子时，X作为叉乘

1. 线的定义

点x_vec = (x, y)'在直线ax + by + c = 0上

标记L = (a, b, c)'，点为(x, y, 1)'为齐次坐标

直线方程可以写成

(x, y, 1)(a, b, c)' = 0或x' * L = 0

2. 直线交点

我们有两条直线L1=(a1, b1, c1)'， L2=(a2, b2, c2)'

假设x是一个点，一个向量，x = L1 x L2 (L1叉乘L2)。

注意，x始终是一个二维点，如果你对(L1xL2)是一个三元素向量，x是一个二维坐标感到困惑，请阅读齐次坐标。

根据三重积，我们知道

L1 * (L1 x L2) = 0, L2 * (L1 x L2) = 0，因为L1,L2共平面

我们用向量x代替L1*x，那么L1*x=0, L2*x=0，这意味着x在L1和L2上，x是交点。

注意，这里x是齐次坐标，如果x的最后一个元素是零，这意味着L1和L2是平行的。

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

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;
}