我添加了第二个答案,因为这与第一个答案有很大不同,试图解决问题中提出的2D装箱算法的更核心问题。使用特定算法时,splitNode
例程生成可用于放置块的down
和right
节点,但并未考虑相邻节点的联合可能容纳更大的块...
例如,在下面的建议算法中,给定一个800x600的初始堆,放置一个500x300的块将导致堆包含两个(0,300)-(800,600)和(500,0)-(800,600)的堆块。这两个堆块将在(500,300)-(800,600)的区域重叠,以确保在寻找适合块的位置时表示最大堆块区域。而在2D装箱算法中,down
或right
中没有交叉区域,而是倾向于潜在重叠空间到一个节点或另一个节点...
以下算法尝试通过实现一组可用的堆块数组来解决这个问题,表示最大可用的块,即使这些堆块彼此重叠。缺点是这引入了一个O(n^2)的算法(unionAll
)来管理堆,加上O(n)的循环(fit
)通过块以适合。因此,这个算法可能接近于O(n^3)的性能,虽然这可能是一个更糟的情况...
Packer = function(w, h) {
this.init(w, h);
};
Packer.prototype = {
init: function(w, h) {
this._root = { x: 0, y: 0, w: w, h: h }
},
intersect: function(block0, block1) {
//
// Returns the intersecting block of
// block0 and block1.
//
let ix0 = Math.max(block0.x0, block1.x0);
let ix1 = Math.min(block0.x1, block1.x1);
let iy0 = Math.max(block0.y0, block1.y0);
let iy1 = Math.min(block0.y1, block1.y1);
if (ix0 <= ix1 && iy0 <= iy1) {
return {x0: ix0, y0: iy0, x1: ix1, y1: iy1};
} else {
return null;
}
},
chunkContains: function(heapBlock0, heapBlock1) {
//
// Determine whether heapBlock0 totally encompasses (ie, contains) heapBlock1.
//
return heapBlock0.x0 <= heapBlock1.x0 && heapBlock0.y0 <= heapBlock1.y0 && heapBlock1.x1 <= heapBlock0.x1 && heapBlock1.y1 <= heapBlock0.y1;
},
expand: function(heapBlock0, heapBlock1) {
//
// Extend heapBlock0 and heapBlock1 if they are
// adjoining or overlapping.
//
if (heapBlock0.x0 <= heapBlock1.x0 && heapBlock1.x1 <= heapBlock0.x1 && heapBlock1.y0 <= heapBlock0.y1) {
heapBlock1.y0 = Math.min(heapBlock0.y0, heapBlock1.y0);
heapBlock1.y1 = Math.max(heapBlock0.y1, heapBlock1.y1);
}
if (heapBlock0.y0 <= heapBlock1.y0 && heapBlock1.y1 <= heapBlock0.y1 && heapBlock1.x0 <= heapBlock0.x1) {
heapBlock1.x0 = Math.min(heapBlock0.x0, heapBlock1.x0);
heapBlock1.x1 = Math.max(heapBlock0.x1, heapBlock1.x1);
}
},
unionMax: function(heapBlock0, heapBlock1) {
//
// Given two heap blocks, determine whether...
//
if (heapBlock0 && heapBlock1) {
// ...heapBlock0 and heapBlock1 intersect, and if so...
let i = this.intersect(heapBlock0, heapBlock1);
if (i) {
if (this.chunkContains(heapBlock0, heapBlock1)) {
// ...if heapBlock1 is contained by heapBlock0...
heapBlock1 = null;
} else if (this.chunkContains(heapBlock1, heapBlock0)) {
// ...or if heapBlock0 is contained by heapBlock1...
heapBlock0 = null;
} else {
// ...otherwise, let's expand both heapBlock0 and
// heapBlock1 to encompass as much of the intersected
// space as possible. In this instance, both heapBlock0
// and heapBlock1 will overlap.
this.expand(heapBlock0, heapBlock1);
this.expand(heapBlock1, heapBlock0);
}
}
}
},
unionAll: function() {
//
// Loop through the entire heap, looking to eliminate duplicative
// heapBlocks, and to extend adjoining or intersecting heapBlocks,
// despite this introducing overlapping heapBlocks.
//
for (let i = 0; i < this.heap.length; i++) {
for (let j = 0; j < this.heap.length; j++) {
if (i !== j) {
this.unionMax(this.heap[i],this.heap[j]);
if (this.heap[i] && this.heap[j]) {
if (this.chunkContains(this.heap[j], this.heap[i])) {
this.heap[i] = null;
} else if (this.chunkContains(this.heap[i], this.heap[j])) {
this.heap[j] = null;
}
}
}
}
}
// Eliminate the duplicative (ie, nulled) heapBlocks.
let onlyBlocks = [];
for (let i = 0; i < this.heap.length; i++) {
if (this.heap[i]) {
onlyBlocks.push(this.heap[i]);
}
}
this.heap = onlyBlocks;
},
fit: function(blocks) {
//
// Loop through all the blocks, looking for a heapBlock
// that it can fit into.
//
this.heap = [{x0:0,y0:0,x1:this._root.w, y1: this._root.h}];
var n, node, block;
for (n = 0; n < blocks.length; n++) {
block = blocks[n];
block.rotate = false;
if (this.findInHeap(block)) {
this.adjustHeap(block);
} else {
// If the block didn't fit in its current orientation,
// rotate its dimensions and look again.
block.w = block.h + (block.h = block.w, 0);
block.rotate = true;
if (this.findInHeap(block)) {
this.adjustHeap(block);
}
}
}
},
findInHeap: function(block) {
//
// Find a heapBlock that can contain the block.
//
for (let i = 0; i < this.heap.length; i++) {
let heapBlock = this.heap[i];
if (heapBlock && block.w <= heapBlock.x1 - heapBlock.x0 && block.h <= heapBlock.y1 - heapBlock.y0) {
block.x0 = heapBlock.x0;
block.y0 = heapBlock.y0;
block.x1 = heapBlock.x0 + block.w;
block.y1 = heapBlock.y0 + block.h;
return true;
}
}
return false;
},
adjustHeap: function(block) {
//
// Find all heap entries that intersect with block,
// and adjust the heap by breaking up the heapBlock
// into the possible 4 blocks that remain after
// removing the intersecting portion.
//
let n = this.heap.length;
for (let i = 0; i < n; i++) {
let heapBlock = this.heap[i];
let overlap = this.intersect(heapBlock, block);
if (overlap) {
// Top
if (overlap.y1 !== heapBlock.y1) {
this.heap.push({
x0: heapBlock.x0,
y0: overlap.y1,
x1: heapBlock.x1,
y1: heapBlock.y1
});
}
// Right
if (overlap.x1 !== heapBlock.x1) {
this.heap.push({
x0: overlap.x1,
y0: heapBlock.y0,
x1: heapBlock.x1,
y1: heapBlock.y1
});
}
// Bottom
if (heapBlock.y0 !== overlap.y0) {
this.heap.push({
x0: heapBlock.x0,
y0: heapBlock.y0,
x1: heapBlock.x1,
y1: overlap.y0
});
}
// Left
if (heapBlock.x0 != overlap.x0) {
this.heap.push({
x0: heapBlock.x0,
y0: heapBlock.y0,
x1: overlap.x0,
y1: heapBlock.y1
});
}
this.heap[i] = null;
}
}
this.unionAll();
}
}
fit
算法将会把结果留在
heap
和传入的
blocks
数组中。例如...
p = new Packer(2400,1200)
blocks = [{w:2100,h:600},{w:2100,h:600},{w:150,h:200},{w:740,h:200},{w:500,h:100}]
p.fit(blocks)
...将使 p.heap
和 blocks
的值如下...
The final HEAP
[{"x0":2100,"y0":940,"x1":2400,"y1":1200},
{"x0":2300,"y0":500,"x1":2400,"y1":1200},
{"x0":2250,"y0":0,"x1":2300,"y1":200}]
The final BLOCKS
[{"w":2100,"h":600,"rotate":false,"x0":0,"y0":0,"x1":2100,"y1":600},
{"w":2100,"h":600,"rotate":false,"x0":0,"y0":600,"x1":2100,"y1":1200},
{"w":150,"h":200,"rotate":false,"x0":2100,"y0":0,"x1":2250,"y1":200},
{"w":200,"h":740,"rotate":true,"x0":2100,"y0":200,"x1":2300,"y1":940},
{"w":100,"h":500,"rotate":true,"x0":2300,"y0":0,"x1":2400,"y1":500}]
请注意,此算法未做优化。即,它没有对输入的块进行排序(例如按宽度、高度或面积等),也没有在执行
unionAll
后对堆进行排序,这会将堆缩减为最大大小的堆块列表。(即,在每次调用
unionAll
后,都有机会按宽度、高度或面积等对堆进行排序,以便在搜索下一个可用的块放置时,如果堆按从大到小排序,则算法将在最大可用堆块中放置块;如果按照从小到大排序,则块将被放置在刚好足够大的堆块中...)无论如何,我们将把这些类型的优化留给读者自己练习。
另外,请以一定的怀疑态度查看此算法,并适当测试。
fit: {x: 2100, y: 600, w: 300, h: 600, used: true}
,因为你应该能够通过循环遍历新提出的算法的结果来构建这个属性。但如果你指的是“down”和“right”属性,这将引入一些将新提出的算法的结果逆向工程到旧格式中的代码,这会增加不必要的复杂性。也就是说,更好的做法是花费时间来改变调用程序以处理新提出的算法的结果。 - Trentium