Geeks 面试题： Box Stacking Problem

Dynamic Programming | Set 22 (Box Stacking Problem)

You are given a set of n types of rectangular 3-D boxes, where the i^th box has height h(i), width w(i) and depth d(i) (all real numbers). You want to create a stack of boxes which is as tall as possible, but you can only stack a box on top of another box if the dimensions of the 2-D base of the lower box are each strictly larger than those of the 2-D base of the higher box. Of course, you can rotate a box so that any side functions as its base. It is also allowable to use multiple instances of the same type of box.

Source:http://people.csail.mit.edu/bdean/6.046/dp/. The link also has video for explanation of solution.

TheBox Stacking problem is a variation of LIS problem. We need to build a maximum height stack.

Following are the key points to note in the problem statement:
1) A box can be placed on top of another box only if both width and depth of the upper placed box are smaller than width and depth of the lower box respectively.
2) We can rotate boxes. For example, if there is a box with dimensions {1x2x3} where 1 is height, 2×3 is base, then there can be three possibilities, {1x2x3}, {2x1x3} and {3x1x2}.
3) We can use multiple instances of boxes. What it means is, we can have two different rotations of a box as part of our maximum height stack.

Following is thesolutionbased onDP solution of LIS problem.

1)Generate all 3 rotations of all boxes. The size of rotation array becomes 3 times the size of original array. For simplicity, we consider depth as always smaller than or equal to width.

2)Sort the above generated 3n boxes in decreasing order of base area.

3)After sorting the boxes, the problem is same as LIS with following optimal substructure property.
MSH(i) = Maximum possible Stack Height with box i at top of stack
MSH(i) = { Max ( MSH(j) ) + height(i) } where j < i and width(j) > width(i) and depth(j) > depth(i).
If there is no such j then MSH(i) = height(i)

4)To get overall maximum height, we return max(MSH(i)) where 0 < i < n

本题理解题意困难，由理解题意到构造算法这个过程原来也不容易。

算法就是最长递增序列算法的灵活运用。

少了原题中的w <= d这个条件，其实也只需要修改一点代码就可以了。

struct Box
{
	int h, w, d;//Here, we don't assume the original Box always keep w <= d, too.
};
int compare(const void *a, const void *b)
{
	Box a1 = (*(Box *)a);
	Box b1 = (*(Box *)b);
	return b1.w * b1.d - a1.w * a1.d;//利用qsort由大到小排序
}
/* Returns the height of the tallest stack that can be formed with give type of boxes */
int maxStackHeight( Box arr[], int n )
{
	Box *narr = new Box[3*n];
	int nLen = 0;
	for (int i = 0; i < n; i++)
	{
		narr[nLen++] = arr[i];
		if (arr[i].w > arr[i].d)//we don't assume w<=d here 
		{
			narr[nLen-1].d = arr[i].w;
			narr[nLen-1].w = arr[i].d;
		}

		narr[nLen].h = arr[i].d;
		narr[nLen].w = arr[i].h < arr[i].w? arr[i].h:arr[i].w;
		narr[nLen++].d = arr[i].h > arr[i].w? arr[i].h:arr[i].w;

		narr[nLen].h = arr[i].w;
		narr[nLen].w = arr[i].d < arr[i].h? arr[i].d:arr[i].h;
		narr[nLen++].d = arr[i].d > arr[i].h? arr[i].d:arr[i].h;
	}
	qsort(narr, nLen, sizeof(Box), compare);
	int *height = new int[nLen];
	for (int i = 0; i < nLen; i++) height[i] = narr[i].h;

	int max_height = 0;
	for (int i = 0; i < nLen; i++)
	{
		for (int j = 0; j < i; j++)
		{
			if (narr[j].w > narr[i].w && narr[j].d > narr[i].d
				&& height[i] < height[j]+narr[i].h)
				height[i] = height[j]+narr[i].h;
			if (max_height < height[i]) max_height = height[i];
		}
	}
	return max_height;
}