diff --git a/480.cpp b/480.cpp
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--- /dev/null
+++ b/480.cpp
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+/* 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;
+}