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形状内文本环绕的算法

[英]Algorithm for Text Wrapping Within a Shape

I am looking for an algorithm to wrap text within a non-rectangular shape, preferably based on the Knuth and Plass algorithm. 我正在寻找一种算法来将文本包装在非矩形形状内,最好是基于Knuth和Plass算法。 The hardest part of this is that the lines may have different heights due to differing font sizes in the text. 最难的部分是由于文本中的字体大小不同,线条可能具有不同的高度。 The image below is an example of what the algorithm should be able to generate. 下图是算法应该能够生成的示例。

形状像心脏的文字

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Try 尝试

text / html text / html

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js JS

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 var text = "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Vivamus blandit nisl eu posuere suscipit. Etiam at quam sed nulla consequat finibus et eget ligula. Nam sit amet imperdiet eros. Ut a congue nibh. Sed ac arcu non risus commodo lobortis et at lorem. Pellentesque pulvinar venenatis pellentesque. Praesent sed pulvinar justo. Ut nec turpis lectus. Suspendisse porta ipsum orci, nec vestibulum tellus luctus quis. Morbi eleifend vel nibh sed rutrum. Etiam feugiat, nunc et efficitur accumsan, quam magna lacinia neque, eu ullamcorper purus turpis eget urna. Etiam lacus mi, gravida vel mollis ut, viverra sed ipsum. Mauris in augue turpis." + "Nulla facilisi. In tristique tortor sit amet leo cursus, sit amet varius ligula sollicitudin. Mauris commodo et sapien id scelerisque. Integer et diam eget arcu bibendum ornare a in tortor. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aliquam vehicula diam ac laoreet tincidunt. Integer sed tellus sit amet lacus bibendum tristique aa nulla. Quisque venenatis suscipit est eget ultricies. Quisque ac orci convallis, interdum enim quis, dictum justo. Phasellus tincidunt nisi vitae justo fringilla, sit amet pellentesque enim dictum. Duis porttitor convallis feugiat. Sed lectus massa, consectetur et vestibulum a, egestas ac eros. Proin mollis ligula tellus, a dignissim eros placerat sed. Donec vel blandit magna. Etiam quis tortor rhoncus, porta ante et, pretium mi." + "Vestibulum ut risus convallis, sagittis orci sed, faucibus erat. Aliquam maximus suscipit lectus, vel pellentesque enim dictum nec. Sed et ex id mauris finibus molestie. Aliquam erat volutpat. Etiam nec ultricies est. Aenean vitae mi id neque vulputate suscipit vel vitae ante. Nam sit amet mi diam. Phasellus efficitur faucibus viverra. Proin cursus purus et tortor dictum elementum. Nunc vehicula erat volutpat, rutrum metus bibendum, blandit sem." + "Nam eu ante eros. Sed pellentesque accumsan neque, eu vulputate lectus luctus ac. In hac habitasse platea dictumst. Nulla vitae dictum diam. Sed eget mauris in felis luctus luctus. Mauris volutpat lacus in est dignissim, sit amet venenatis ipsum laoreet. Phasellus semper id velit ut tristique. Proin dictum erat at mauris lacinia blandit. Ut scelerisque mi ex, ut pretium elit commodo sit amet. Praesent faucibus turpis tellus, ac vulputate felis aliquet eget. Ut est urna, dignissim nec risus sit amet, eleifend lacinia enim. Donec non massa in purus efficitur placerat. Donec accumsan neque elit, eget vulputate metus finibus quis. Ut interdum sem vel felis imperdiet pretium. Integer erat neque, bibendum id egestas nec, facilisis sed massa." + "Curabitur sit amet mattis odio. Mauris mattis turpis eu lectus eleifend, semper ultrices elit malesuada. Mauris fringilla arcu sed sem vehicula vulputate. Donec semper nisl ut urna bibendum, quis eleifend tortor commodo. Pellentesque vehicula est et neque vulputate accumsan. Duis faucibus sodales lacus et suscipit. Pellentesque tortor enim, pulvinar scelerisque pretium non, aliquam sed risus. Vestibulum aliquam dolor ipsum, eget pretium ante fermentum a. Donec semper lectus pulvinar tellus hendrerit sagittis. Nunc blandit sed ligula vel consequat. Etiam sed est quis nulla blandit blandit. Sed in enim pulvinar elit accumsan faucibus sed sodales lorem. Aenean ac efficitur arcu, sed fringilla lectus." , i = -1; var elem = document.querySelector("pre"); elem.innerText = elem.innerText.replace(/[^\\n|\\r|\\t|\\s+]/g, function() { return text[++i] }); 
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For a simple algorithm, let's assume that you can compute a bounding box around each word, and that you have an image with a mask of the shape that you want to fill. 对于一个简单的算法,我们假设您可以计算每个单词周围的边界框,并且您有一个图像,其中包含您要填充的形状的蒙版。

Sweep down from the top of the image mask until you find a line that is as long as the first word. 从图像蒙版的顶部向下扫描,直到找到与第一个单词一样长的行。 See if you can extend it downward into a rectangle the size of the bounding box. 看看是否可以将其向下延伸到边界框大小的矩形中。 If so, drop the first word there. 如果是这样,请将第一个单词放在那里。 If not, keep sweeping. 如果没有,继续扫地。

Once you drop a word, see if you can extend the bounding box to be the width of (first box + second box + space) and the height of max(first box,second box). 删除单词后,查看是否可以将边界框扩展为(第一个框+第二个框+空格)的宽度和最大高度(第一个框,第二个框)。 If so, drop the second word there. 如果是这样,请将第二个字放在那里。 If not, center the first word left-to-right within the bounding box that will fit within your image mask (left to right), remove that bounding box from the mask, and keep going. 如果没有,请将第一个单词从左到右居中放置在适合图像蒙版(从左到右)的边界框内,从蒙版中移除该边界框,然后继续。

You can make this slightly fancier by insisting that lines have the same baseline even if broken by the shape (eg lines across the top nubs of the heart); 你可以通过坚持线条具有相同的基线即使被形状打破(例如,心脏的顶部小块上的线条),也可以使它略微变得更加迷人; you then need to have an alternate "continue along this baseline" condition. 然后你需要有一个替代“沿着这个基线继续”的条件。 But the basic idea above, with an image mask that you use to try to fit rectangles inside which are removed when you're done, will do the job. 但是上面的基本思想,使用一个图像蒙板,你用来试图在你完成之后去掉那些被删除的矩形,它将完成这项工作。

(It is faster to use geometric operations than pixel-based ones described here, but then one has to worry about all the cases for figuring out how a bounding box fits within an arbitrary polygon, and that's a bit long to explain here.) (使用几何操作比使用此处描述的基于像素的操作更快,但是人们必须担心所有的情况,以确定边界框如何适合任意多边形,这在这里解释有点长。)

Let's say each letter has a specific size(width and height, in this case we probably only care about width because all letters have the same height). 假设每个字母都有一个特定的大小(宽度和高度,在这种情况下,我们可能只关心宽度,因为所有字母都有相同的高度)。 Then we need the following: 然后我们需要以下内容:

  1. Wrapped objects of each word - to make sure there's no fragmentation of the word 每个单词的被包裹的对象 - 以确保单词没有碎片
  2. given width of each word. 给定每个单词的宽度。
  3. Partition the polygon into chunks that are the heights of the word. 将多边形划分为块的高度。 So you have x horizontal strips that make up the image. 所以你有x个水平条纹组成图像。
  4. Then, find the width of the inside polygon if you were to inscribe words inside. 然后,如果要在里面刻字,找到内部多边形的宽度。 This means you're removing the round edges of the heart. 这意味着你要去除心脏的圆形边缘。 Where a vertical line can intersect the edge of the heart on the strip is where you remove the round edge. 如果垂直线可以与条带上的心脏边缘相交,则可以移除圆边。

Now that we have the size of each strip of the image, let's say we have this (i'm using arbitary unit for width): 既然我们有图像的每个条带的大小,让我们说我们有这个(我使用宽度的任意单位):

[_________] [__________]  <-- 10 width (5 each)
  [__________________] <-- 9 width
    [_____________] <-- 7 width   
       [_______] <-- 5 width
         [___] <-- 3 width
          [_] <-- 2 width

EDIT: Just realized how ugly the heart is, my bad. 编辑:刚刚意识到心脏是多么丑陋,我的坏。

Now we need the size of each word, and insert them sequentially. 现在我们需要每个单词的大小,然后按顺序插入它们。 Each block has x width, and each word object has y width. 每个块具有x宽度,每个字对象具有y宽度。 If y width > x width, we move to the next line and check there 如果y width> x width,我们移动到下一行并检查那里

considering we already have widths
while(image.hasNextChunk()){
    currentChunk = image.nextChunk();
    if(currentWord.width < currentChunk.width) //insert here and then change currentWord to nextWord
    ...
}

I think this is what you want, but I'm not completely sure. 我想这就是你想要的,但我不完全确定。 Let me know if this helped! 如果这有帮助,请告诉我! :) :)

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