问题
I'm trying to find an algorithm to find K disjoint, contiguous subsets of size L of an array x of real numbers that maximize the sum of the elements.
Spelling out the details, X is a set of N positive real numbers:X={x[1],x[2],...x[N]} where x[j]>=0 for all j=1,...,N.
A contiguous subset of length L called S[i] is defined as L consecutive members of X starting at position n[i] and ending at position n[i]+L-1:S[i] = {x[j] | j=n[i],n[i]+1,...,n[i]+L-1} = {x[n[i]],x[n[i]+1],...,x[n[i]+L-1]}.
Two of such subsets S[i] and S[j] are called pairwise disjoint (non-overlapping) if |n[i]-n[j]|>=L. In other words, they don't contain any identical members of X.
Define the summation of the members of each subset:
SUM[i] = x[n[i]]+x[n[i]+1]+...+x[n[i]+L-1];
The goal is find K contiguous and disjoint(non-overlapping) subsets S[1],S[2],...,S[K] of length L such that SUM[1]+SUM[2]+...+SUM[K] is maximized.
回答1:
This is solved by dynamic programming. Let M[i] be the best solution only for the first i elements of x. Then:
M[i] = 0 for i < L
M[i] = max(M[i-1], M[i-L] + sum(x[i-L+1] + x[i-L+2] + ... + x[i]))
The solution to your problem is M[N].
When you code it, you can incrementally compute the sum (or simply pre-compute all the sums) leading to an O(N) solution in both space and time.
If you have to find exactly K subsets, you can extend this, by defining M[i, k] to be the optimal solution with k subsets on the first i elements. Then:
M[i, k] = 0 for i < k * L or k = 0.
M[i, k] = max(M[i-1, k], M[i-L, k-1] + sum(x[i-L+1] + ... + x[i])
The solution to your problem is M[N, K].
This is a 2d dynamic programming solution, and has time and space complexity of O(NK) (assuming you use the same trick as above for avoiding re-computing the sum).
来源：https://stackoverflow.com/questions/29268442/maximizing-the-overall-sum-of-k-disjoint-and-contiguous-subsets-of-size-l-among