学步园

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.

The Box 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 the solution based on DP 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

3维的就确定下两维（底面积），根据底面积排序，然后找高度的LDS

package DP;

import java.util.Arrays;
import java.util.Comparator;

public class BoxStacking {

	public static void main(String[] args) {
		Box[] A = new Box[4];
		A[0] = new Box(4,6,7);
		A[1] = new Box(1,2,3);
		A[2] = new Box(4,5,6);
		A[3] = new Box(10,12,32);
		int n = A.length;
		System.out.println(maxStackHeight(A, n));
	}	
	
	// Time Complexity: O(n^2) Auxiliary Space: O(n)
	public static int maxStackHeight(Box[] A, int n){
		/* Create an array of all rotations of given boxes
	      For example, for a box {1, 2, 3}, we consider three
	      instances{{1, 2, 3}, {2, 1, 3}, {3, 1, 2}} */
		Box[] rot = new Box[3*n];
		int index = 0;
		for(int i=0; i<n; i++){
			rot[index] = A[i];		// Copy the original box
			index++;
			
			rot[index] = new Box();		// First rotation of box
			rot[index].h = A[i].w;
			rot[index].d = Math.max(A[i].h, A[i].d);
			rot[index].w = Math.min(A[i].h, A[i].d);
			index++;
			
			rot[index] = new Box();		// Second rotation of box
			rot[index].h = A[i].d;
			rot[index].d = Math.max(A[i].h, A[i].w);
			rot[index].w = Math.min(A[i].h, A[i].w);
			index++;
		}
		
		n = 3*n;		// Now the number of boxes is 3n
		
		/* Sort the array ‘rot[]‘ in decreasing order, using library
	      function for quick sort */
		Arrays.sort(rot, new Comparator<Box>() {
			@Override
			public int compare(Box o1, Box o2) {		// 按底面积递减排序
				return (o2.d*o2.w) - (o1.d*o1.w);
			}
		});
		
		/* Initialize msh values for all indexes 
	      msh[i] –> Maximum possible Stack Height with box i on top */
		int[] msh = new int[n];
		for(int i=0; i<n; i++){
			msh[i] = rot[i].h;
		}
		
		/* Compute optimized msh values in bottom up manner */
		for(int i=1; i<n; i++){
			for(int j=0; j<i; j++){
				if(rot[i].w<rot[j].w && rot[i].d<rot[j].d){
					msh[i] = Math.max(msh[i], msh[j]+rot[i].h);
				}
			}
		}
		
		/* Pick maximum of all msh values */
		int max = 0;
		for(int i=0; i<n; i++){
			max = Math.max(max, msh[i]);
		}
		return max;
	}

	public static class Box{
		int h, w, d;		// h –> height, w –> width, d –> depth
		// for simplicity of solution, always keep w <= d
		public Box(){
		}
		public Box(int h_, int w_, int d_){
			h = h_;
			w = w_;
			d = d_;
		}
	}
}

http://www.geeksforgeeks.org/dynamic-programming-set-21-box-stacking-problem/

http://people.csail.mit.edu/bdean/6.046/dp/