我正在尝试使用最大矩形算法实现2D bin打包,如下文所述。
http://clb.demon.fi/files/RectangleBinPack.pdf
为了实现这一点,哪种类型的数据结构最合适? 谷歌搜索后,我发现使用树木的断头台打包算法有不同的实现。也可以采用相同的方法。 算法本身对我来说不太清楚。我是否可以对此有更多的澄清。
答案 0 :(得分:1)
在研究自动CSS精灵生成时,我偶然发现了一些东西。我认为在源代码中由元组表示的矩形,实际算法使用二叉树。也许它对你有用。
http://codeincomplete.com/posts/2011/5/7/bin_packing/
修改强> 这是https://github.com/jakesgordon/bin-packing/blob/master/js/packer.js
的代码示例/******************************************************************************
This is a very simple binary tree based bin packing algorithm that is initialized
with a fixed width and height and will fit each block into the first node where
it fits and then split that node into 2 parts (down and right) to track the
remaining whitespace.
Best results occur when the input blocks are sorted by height, or even better
when sorted by max(width,height).
Inputs:
------
w: width of target rectangle
h: height of target rectangle
blocks: array of any objects that have .w and .h attributes
Outputs:
-------
marks each block that fits with a .fit attribute pointing to a
node with .x and .y coordinates
Example:
-------
var blocks = [
{ w: 100, h: 100 },
{ w: 100, h: 100 },
{ w: 80, h: 80 },
{ w: 80, h: 80 },
etc
etc
];
var packer = new Packer(500, 500);
packer.fit(blocks);
for(var n = 0 ; n < blocks.length ; n++) {
var block = blocks[n];
if (block.fit) {
Draw(block.fit.x, block.fit.y, block.w, block.h);
}
}
******************************************************************************/
Packer = function(w, h) {
this.init(w, h);
};
Packer.prototype = {
init: function(w, h) {
this.root = { x: 0, y: 0, w: w, h: h };
},
fit: function(blocks) {
var n, node, block;
for (n = 0; n < blocks.length; n++) {
block = blocks[n];
if (node = this.findNode(this.root, block.w, block.h))
block.fit = this.splitNode(node, block.w, block.h);
}
},
findNode: function(root, w, h) {
if (root.used)
return this.findNode(root.right, w, h) || this.findNode(root.down, w, h);
else if ((w <= root.w) && (h <= root.h))
return root;
else
return null;
},
splitNode: function(node, w, h) {
node.used = true;
node.down = { x: node.x, y: node.y + h, w: node.w, h: node.h - h };
node.right = { x: node.x + w, y: node.y, w: node.w - w, h: h };
return node;
}
}