Box Stacking Problem.cpp

#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; 
}