Box Stacking Problem

/* Dynamic Programming implementation of Box Stacking problem */
#include<stdio.h> 
#include<stdlib.h> 

/* Representation of a box */
struct Box 
{ 
// h --> height, w --> width, d --> depth 
int h, w, d; // for simplicity of solution, always keep w <= d 
}; 

// A utility function to get minimum of two intgers 
int min (int x, int y) 
{ return (x < y)? x : y; } 

// A utility function to get maximum of two intgers 
int max (int x, int y) 
{ return (x > y)? x : y; } 

/* Following function is needed for library function qsort(). We 
use qsort() to sort boxes in decreasing order of base area. 
Refer following link for help of qsort() and compare() 
http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
int compare (const void *a, const void * b) 
{ 
	return ( (*(Box *)b).d * (*(Box *)b).w ) - 
		( (*(Box *)a).d * (*(Box *)a).w ); 
} 

/* Returns the height of the tallest stack that can be 
formed with give type of boxes */
int maxStackHeight( Box arr[], 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[3*n]; 
int index = 0; 
for (int i = 0; i < n; i++) 
{ 
	// Copy the original box 
	rot[index].h = arr[i].h; 
	rot[index].d = max(arr[i].d, arr[i].w); 
	rot[index].w = min(arr[i].d, arr[i].w); 
	index++; 

	// First rotation of box 
	rot[index].h = arr[i].w; 
	rot[index].d = max(arr[i].h, arr[i].d); 
	rot[index].w = min(arr[i].h, arr[i].d); 
	index++; 

	// Second rotation of box 
	rot[index].h = arr[i].d; 
	rot[index].d = max(arr[i].h, arr[i].w); 
	rot[index].w = min(arr[i].h, arr[i].w); 
	index++; 
} 

// Now the number of boxes is 3n 
n = 3*n; 

/* Sort the array 'rot[]' in non-increasing order 
	of base area */
qsort (rot, n, sizeof(rot[0]), compare); 

// Uncomment following two lines to print all rotations 
// for (int i = 0; i < n; i++ ) 
// printf("%d x %d x %d\n", rot[i].h, rot[i].w, rot[i].d); 

/* Initialize msh values for all indexes 
	msh[i] --> Maximum possible Stack Height with box i on top */
int msh[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] < msh[j] + rot[i].h 
			) 
		{ 
			msh[i] = msh[j] + rot[i].h; 
		} 


/* Pick maximum of all msh values */
int max = -1; 
for ( int i = 0; i < n; i++ ) 
	if ( max < msh[i] ) 
		max = msh[i]; 

return max; 
} 

/* Driver program to test above function */
int main() 
{ 
Box arr[] = { {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32} }; 
int n = sizeof(arr)/sizeof(arr[0]); 

printf("The maximum possible height of stack is %d\n", 
		maxStackHeight (arr, n) ); 

return 0; 
}