Copying Books UVa Online Judge Dynamic Programing Solution - algorithm

Question

I can solve Copying Books Problem using binary search method as it is easy to implement. But I have just started solving Dynamic Programing problems and I wanted to know Dynamic Programing solution for the problem
Before the invention of book-printing, it was very hard to make a
copy of a book. All the contents had to be re-written by hand by so
called scribers. The scriber had been given a book and after several
months he finished its copy. One of the most famous scribers lived in
the 15th century and his name was Xaverius Endricus Remius Ontius
Xendrianus (Xerox). Anyway, the work was very annoying and boring. And
the only way to speed it up was to hire more scribers.
Once upon a time, there was a theater ensemble that wanted to play
famous Antique Tragedies. The scripts of these plays were divided into
many books and actors needed more copies of them, of course. So they
hired many scribers to make copies of these books. Imagine you have m
books (numbered 1, 2, ...., m) that may have different number of
pages ( p_1, p_2, ..., p_m) and you want to make one copy of each
of them. Your task is to divide these books among k scribes, k <=
m. Each book can be assigned to a single scriber only, and every
scriber must get a continuous sequence of books. That means, there
exists an increasing succession of numbers 0 = b_0 < b_1 < b_2, ...
< b_{k-1} <= b_k = m$ such that i-th scriber gets a sequence of books
with numbers between bi-1+1 and bi. The time needed to make a copy of
all the books is determined by the scriber who was assigned the most
work. Therefore, our goal is to minimize the maximum number of pages
assigned to a single scriber. Your task is to find the optimal
assignment.
For Binary Search I am doing the following.
Low =1 and High = Sum of pages of all books
Run Binary search
For Mid(Max pages assigned to a scribe), assign books greedily such that no scribe gets page more than MAX
If scribes remain without work it means actual value is less than MID, if Books remain actual pages is more MID and I am updating accordingly.

Answer 1

Here is a possible dynamic programming solution written in python. I use indexing starting from 0.
k = 2 # number of scribes
# number of pages per book. 11 pages for first book, 1 for second, etc.
pages = [11, 1, 1, 10, 1, 1, 3, 3]
m = len(pages) # number of books
def find_score(assignment):
max_pages = -1
for scribe in assignment:
max_pages = max(max_pages, sum([pages[book] for book in scribe]))
return max_pages
def find_assignment(assignment, scribe, book):
if book == m:
return find_score(assignment), assignment
assign_current = [x[:] for x in assignment] # deep copy
assign_current[scribe].append(book)
current = find_assignment(assign_current, scribe, book + 1)
if scribe == k - 1:
return current
assign_next = [x[:] for x in assignment] # deep copy
assign_next[scribe + 1].append(book)
next = find_assignment(assign_next, scribe + 1, book + 1)
return min(current, next)
initial_assignment = [[] for x in range(k)]
print find_assignment(initial_assignment, 0, 0)
The function find_assignment returns the assignment as a list where the ith element is a list of book indexes assigned to the ith scribe. The score of the assignment is returned as well (the max number of pages a scribe has to copy in the assignment).
The key to dynamic programming is to first identify the subproblem. In this case, the books are ordered and can only be assigned sequentially. Thus the subproblem is to find an optimal assignment for the last n books using s scribes (where n < m and s < k). A subproblem can be solved with smaller subproblems using the following relation: min(assigning book to the "current" scribe, assigning the book to the next scribe).