基于动态规划制表法的工人-任务分配最大收益求解方案咨询

Absolutely—this is a classic resource allocation dynamic programming problem that’s tailor-made for a tabulation approach. Let’s break this down step by step, starting with clearing up that common misconception about subproblems being independent.

Your initial thought that subproblems are fully independent makes sense at first glance, but here’s the key: we have a fixed pool of m workers. The number of workers you allocate to the first j tasks directly limits how many you can allocate to the remaining tasks. This creates a clear dependency between subproblems, which is exactly what dynamic programming leverages.

The right way to frame subproblems here is:

What’s the maximum total gain we can get by assigning t workers to the first j tasks?

Once we define subproblems this way, we can build up solutions to larger problems using smaller ones.

Let’s walk through building the DP table, step by step.

1. Define the DP Table #

We’ll create a 2D array dp where:

• dp[j][t] = maximum total收益 (gain) from assigning t workers to the first j tasks.
• The table dimensions are (n+1) x (m+1) (we include j=0 for "no tasks" and t=0 for "no workers" as base cases).

2. Initialize the Base Cases #

• No tasks: For any number of workers t, dp[0][t] = 0 (there’s no task to assign workers to, so gain is zero).
• No workers: For any number of tasks j, dp[j][0] = sum of A_k[0] for k=1 to j (assigning 0 workers to each task gives the sum of each task’s 0-worker gain; often A_k[0] = 0, so this simplifies to 0).

3. State Transition Logic #

For each task j (from 1 to n), and each possible worker count t (from 1 to m), we calculate dp[j][t] by considering every possible number of workers i we could assign to task j (from 0 to t):

dp[j][t] = max( dp[j-1][t - i] + A_j[i] ) for all i in 0 ≤ i ≤ t

What this means:

• dp[j-1][t - i] is the best gain from using t-i workers on the first j-1 tasks.
• A_j[i] is the gain from assigning i workers to task j.
• We take the maximum value across all possible i to get the optimal gain for j tasks and t workers.

Since A_j[i] is non-decreasing, this doesn’t change the logic—we still need to check all possible i because combining a lower i for task j with a higher gain from the first j-1 tasks might yield a better total than a higher i for task j.

4. Example Walkthrough (Matching Your Test Case) #

Let’s use your example where m=3 workers yield a maximum gain of 15 (from assigning 2 workers to task 2 and 1 to task 4), and m=4 workers yield 100 (assigning all 4 to task 1):

1. Fill row j=1 (first task):
   • dp[1][4] = A_1[4] = 100 (this is the value that will drive the m=4 result).
2. Fill row j=2 (second task):
   • For t=2, dp[2][2] = max(dp[1][2]+A_2[0], dp[1][1]+A_2[1], dp[1][0]+A_2[2]). If A_2[2] is the best option here, this becomes A_2[2].
3. Fill rows j=3 and j=4:
   • When we reach j=4 and t=3, we check all possible i (0 to 3) for task 4. The maximum comes from i=1: dp[3][2] + A_4[1] = A_2[2] + A_4[1] = 15.

5. Optional Space Optimization #

If space is a concern, we can reduce the table to a 1D array dp[t] (size m+1). Since we only need the previous row (j-1) to calculate the current row (j), we can update the array from right to left (to avoid overwriting values we still need):

Initialize dp[t] = 0 for all t (assuming A_k[0] = 0)
For each task j from 1 to n:
    For t from m down to 0:
        For i from 1 to t:
            dp[t] = max(dp[t], dp[t - i] + A_j[i])

This cuts the space complexity from O(n*m) to O(m).

• Brute force: O(n^m) (exponential, which becomes infeasible quickly as m grows).
• Tabulation DP: O(n*m²) (polynomial—for most practical values of n and m, this is way more efficient).

Tabulation is not only possible here—it’s the most intuitive way to implement this solution. You’ll systematically fill the table row by row, and each entry builds directly from smaller subproblems, making it easy to verify and debug.

内容的提问来源于stack exchange，提问作者Nova