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javascript - 移动矩形,使它们不重叠

转载 作者:数据小太阳 更新时间:2023-10-29 04:46:44 27 4
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这是一个一半是编程,一半是数学的问题。

我有一些方框,它们表示为四个 Angular 点。它们是真正的矩形,是两组平行线的交点,每组中的每条线都与另一组中的两条线成直 Angular (这样我们就清楚了。)

对于任何一组 n 个框,我如何才能有效地计算将它们移动到哪里(最小距离)以使它们不会彼此重叠?

我在这里使用 javascript 工作。这是数据:

//an array of indefinite length of boxes
//boxes represented as arrays of four points
//points represented as arrays of two things, an x and a y, measured in
//pixels from the upper left corner

var boxes = [[[504.36100124308336,110.58685958804978],[916.3610012430834,110.58685958804978],[916.3610012430834,149.58685958804978],[504.36100124308336,149.58685958804978]],[[504.4114378910622,312.3334473005064],[554.4114378910622,312.3334473005064],[554.4114378910622,396.3334473005064],[504.4114378910622,396.3334473005064]],[[479.4272869132357,343.82042608058134],[516.4272869132358,343.82042608058134],[516.4272869132358,427.82042608058134],[479.4272869132357,427.82042608058134]],[[345.0558946408693,400.12499171846],[632.0558946408694,400.12499171846],[632.0558946408694,439.12499171846],[345.0558946408693,439.12499171846]],[[164.54073131913765,374.02074227992966],[264.54073131913765,374.02074227992966],[264.54073131913765,428.02074227992966],[164.54073131913765,428.02074227992966]],[[89.76601656567325,257.7956256799442],[176.76601656567325,257.7956256799442],[176.76601656567325,311.7956256799442],[89.76601656567325,311.7956256799442]],[[60.711850703535845,103.10558195262593],[185.71185070353584,103.10558195262593],[185.71185070353584,157.10558195262593],[60.711850703535845,157.10558195262593]],[[169.5240557746245,23.743626531766495],[231.5240557746245,23.743626531766495],[231.5240557746245,92.7436265317665],[169.5240557746245,92.7436265317665]],[[241.6776988694169,24.30106373152889],[278.6776988694169,24.30106373152889],[278.6776988694169,63.30106373152889],[241.6776988694169,63.30106373152889]],[[272.7734457459479,15.53275710947554],[305.7734457459479,15.53275710947554],[305.7734457459479,54.53275710947554],[272.7734457459479,54.53275710947554]],[[304.2905062327675,-3.9599943474960035],[341.2905062327675,-3.9599943474960035],[341.2905062327675,50.04000565250399],[304.2905062327675,50.04000565250399]],[[334.86335590542114,12.526345270766143],[367.86335590542114,12.526345270766143],[367.86335590542114,51.52634527076614],[334.86335590542114,51.52634527076614]],[[504.36100124308336,110.58685958804978],[916.3610012430834,110.58685958804978],[916.3610012430834,149.58685958804978],[504.36100124308336,149.58685958804978]],[[504.4114378910622,312.3334473005064],[554.4114378910622,312.3334473005064],[554.4114378910622,396.3334473005064],[504.4114378910622,396.3334473005064]],[[479.4272869132357,343.82042608058134],[516.4272869132358,343.82042608058134],[516.4272869132358,427.82042608058134],[479.4272869132357,427.82042608058134]],[[345.0558946408693,400.12499171846],[632.0558946408694,400.12499171846],[632.0558946408694,439.12499171846],[345.0558946408693,439.12499171846]],[[164.54073131913765,374.02074227992966],[264.54073131913765,374.02074227992966],[264.54073131913765,428.02074227992966],[164.54073131913765,428.02074227992966]],[[89.76601656567325,257.7956256799442],[176.76601656567325,257.7956256799442],[176.76601656567325,311.7956256799442],[89.76601656567325,311.7956256799442]],[[60.711850703535845,103.10558195262593],[185.71185070353584,103.10558195262593],[185.71185070353584,157.10558195262593],[60.711850703535845,157.10558195262593]],[[169.5240557746245,23.743626531766495],[231.5240557746245,23.743626531766495],[231.5240557746245,92.7436265317665],[169.5240557746245,92.7436265317665]],[[241.6776988694169,24.30106373152889],[278.6776988694169,24.30106373152889],[278.6776988694169,63.30106373152889],[241.6776988694169,63.30106373152889]],[[272.7734457459479,15.53275710947554],[305.7734457459479,15.53275710947554],[305.7734457459479,54.53275710947554],[272.7734457459479,54.53275710947554]],[[304.2905062327675,-3.9599943474960035],[341.2905062327675,-3.9599943474960035],[341.2905062327675,50.04000565250399],[304.2905062327675,50.04000565250399]],[[334.86335590542114,12.526345270766143],[367.86335590542114,12.526345270766143],[367.86335590542114,51.52634527076614],[334.86335590542114,51.52634527076614]]]

This fiddle shows the boxes drawn on a canvas semi-transparently for clarity.

最佳答案

您可以使用贪心算法。它远非最佳,但可能“足够好”。这是一个草图:

 1 Sort the rectangles by the x-axis, topmost first. (n log n)
2 for each rectangle r1, top to bottom
//check for intersections with the rectangles below it.
// you only have to check the first few b/c they are sorted
3 for every other rectangle r2 that might intersect with it
4 if r1 and r2 intersect //this part is easy, see @Jose's answer
5 left = the amount needed to resolve the collision by moving r2 left
6 right = the amount needed to resolve the collision by moving r2 right
7 down = the amount needed to resolve the collision by moving r2 down

8 move r2 according to the minimum value of (left, right down)
// (this may create new collisions, they will be resolved in later steps)
9 end if

10 end
11 end

注意 第 8 步可能会与先前的矩形产生新的碰撞,而该碰撞无法正确解决。嗯。您可能需要携带一些关于以前的矩形的元数据来避免这种情况。思考...

关于javascript - 移动矩形,使它们不重叠,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/6750377/

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