两个矩形交叉点

时间:2012-11-15 01:37:29

标签: algorithm math pseudocode shapes

我有两个矩形,每个矩形的特征是4个值:

左侧位置X,顶部位置Y,宽度W和高度H

X1, Y1, H1, W1
X2, Y2, H2, W2

矩形不会旋转,如下所示:

+--------------------> X axis
|
|    (X,Y)      (X+W, Y)
|    +--------------+
|    |              |
|    |              |
|    |              |
|    +--------------+
v    (X, Y+H)     (X+W,Y+H)

Y axis

确定两个矩形的交点是否为空的最佳解决方案是什么?

7 个答案:

答案 0 :(得分:88)

if (X1+W1<X2 or X2+W2<X1 or Y1+H1<Y2 or Y2+H2<Y1):
    Intersection = Empty
else:
    Intersection = Not Empty

如果您有四个坐标 - ((X,Y),(A,B))((X1,Y1),(A1,B1)) - 而不是两个加宽度和高度,它将如下所示:

if (A<X1 or A1<X or B<Y1 or B1<Y):
    Intersection = Empty
else:
    Intersection = Not Empty

答案 1 :(得分:4)

最好的例子..

/**
 * Check if two rectangles collide
 * x_1, y_1, width_1, and height_1 define the boundaries of the first rectangle
 * x_2, y_2, width_2, and height_2 define the boundaries of the second rectangle
 */
boolean rectangle_collision(float x_1, float y_1, float width_1, float height_1, float x_2, float y_2, float width_2, float height_2)
{
  return !(x_1 > x_2+width_2 || x_1+width_1 < x_2 || y_1 > y_2+height_2 || y_1+height_1 < y_2);
}

以及另一种方式看到这个link ...并自己编码......

答案 2 :(得分:2)

如果两个矩形具有相同的尺寸,您可以这样做:

if (abs (x1 - x2) < w && abs (y1 - y2) < h) {
    // overlaps
}

答案 3 :(得分:0)

我刚试过一个c程序,并写在下面。

#include<stdio.h>

int check(int i,int j,int i1,int j1, int a, int b,int a1,int b1){
    return (\
    (((i>a) && (i<a1)) && ((j>b)&&(j<b1))) ||\ 
    (((a>i) && (a<i1)) && ((b>j)&&(b<j1))) ||\ 
    (((i1>a) && (i1<a1)) && ((j1>b)&&(j1<b1))) ||\ 
    (((a1>i) && (a1<i1)) && ((b1>j)&&(b1<j1)))\
    );  
}
int main(){
    printf("intersection test:(0,0,100,100),(10,0,1000,1000) :is %s\n",check(0,0,100,100,10,0,1000,1000)?"intersecting":"Not intersecting");
    printf("intersection test:(0,0,100,100),(101,101,1000,1000) :is %s\n",check(0,0,100,100,101,101,1000,1000)?"intersecting":"Not intersecting");
    return 0;
}

答案 4 :(得分:0)

使用坐标系,其中(0,0)是左上角。

我从垂直和水平滑动窗口的角度来考虑它 并提出这个:

(B.Bottom&gt; A.Top&amp;&amp; B.Top&lt; A.Bottom)&amp;&amp; (B.Right&gt; A.Left&amp;&amp; B.Left&lt; A.Right)

如果您将DeMorgan定律应用于以下内容,您将获得以下内容:

不是(B.Bottom&lt; A.Top || B.Top&gt; A.Bottom || B.Right&lt; A.Left || B.Left&gt; A.Right)

  1. B高于A
  2. B低于A
  3. B留在A
  4. B是A
  5. 的权利

答案 5 :(得分:0)

如果左下角和右上角的矩形坐标为:
rect1和
的(r1x1,r1y1),(r1x2,r1y2) rect2的(r2x1,r2y1),(r2x2,r2y2) (Python如下代码所示)

    intersect = False
    for x in [r1x1, r1x2]:
        if (r2x1<=x<=r2x2):
            for y in [r1y1, r1y2]:
                if (r2y1<=y<=r2y2):
                    intersect = True
                    return intersect
                else:
                    for Y in [r2y1, r2y2]:
                        if (r1y1<=Y<=r1y2):
                            intersect = True
                            return intersect
        else:  
            for X in [r2x1, r2x2]:
                if (r1x1<=X<=r1x2):
                    for y in [r2y1, r2y2]:
                        if (r1y1<=y<=r1y2):
                            intersect = True
                            return intersect
                        else:
                            for Y in [r1y1, r1y2]:
                                if (r2y1<=Y<=r2y2):
                                    intersect = True
                                    return intersect
    return intersect

答案 6 :(得分:-1)

if(X1 <= X2 + W2&amp;&amp; X2&lt; = X1 + W1&amp; Y1&gt; = Y2-H2&amp; Y2&gt; = Y1 + H1)    相交

问题Y是最高位置..

注意:此解决方案仅在矩形与X / Y轴对齐时才有效。