You are working in a warehouse and want to pass the time and make the tallest tower of boxes you can. Your daily shipment of boxes includes two 1-indexed arrays, and .
Array Structure is the array of box names (strings) - relatively unimportant for you, for all you care about is… is the array of weight values - is how heavy the th box will be. The array is ordered by the time in which you'll receive the box. Constraints Later items in the array must be added to your stack later than earlier items - the shipment arrays are ordered lists. You want each box on the stack to be lighter than anything it sits on top of, so you'd like to select a combination of boxes which, when stacked in array order, have strictly decreasing weight value .
The order of the array cannot be changed. Goals Pick a set of boxes so that each box is lighter than the one selected before it. Pick as many boxes as possible. We will solve this with the following overlapping subproblem: Let = {the maximum number of boxes that can be stacked from the first boxes under the restrictions above, that must include using the box in particular}. Example For the following items: "Air Container", "Bulky object", "Cardboard Quadrilateral", "Delicate Item", "Empty", "Fragile Cube", "Gravel" tells us how many items can be stacked from the first 5 boxes, under the constraint that we must select box 5. (Which is 3 boxes - we could stack “Bulky Object”, “Cardboard Quadrilateral” and “Empty”.) Your task Give the recurrence relation for .
Include all cases (and only those) in the expression .
Ensure no cases selected overlap. Assume that the maximum/minimum of an empty set is .
Multiple solutions may be possible, and multiple cases may be unnecessary but not incorrect to include. All correct approaches will achieve full marks.

Question

You are working in a warehouse and want to pass the time and make the tallest tower of boxes you can. Your daily shipment of boxes includes two 1-indexed arrays, and .
Array Structure is the array of box names (strings) - relatively unimportant for you, for all you care about is… is the array of weight values -  is how heavy the th box will be. The array is ordered by the time in which you'll receive the box. Constraints Later items in the array must be added to your stack later than earlier items - the shipment arrays are ordered lists. You want each box on the stack to be lighter than anything it sits on top of, so you'd like to select a combination of boxes which, when stacked in array order, have strictly decreasing weight value .
The order of the array cannot be changed. Goals Pick a set of boxes so that each box is lighter than the one selected before it. Pick as many boxes as possible. We will solve this with the following overlapping subproblem: Let = {the maximum number of boxes that can be stacked from the first boxes under the restrictions above, that must include using the box in particular}. Example For the following items: "Air Container", "Bulky object", "Cardboard Quadrilateral", "Delicate Item", "Empty", "Fragile Cube", "Gravel" tells us how many items can be stacked from the first 5 boxes, under the constraint that we must select box 5. (Which is 3 boxes - we could stack “Bulky Object”, “Cardboard Quadrilateral” and “Empty”.) Your task Give the recurrence relation for .
Include all cases (and only those) in the expression .
Ensure no cases selected overlap. Assume that the maximum/minimum of an empty set is .
Multiple solutions may be possible, and multiple cases may be unnecessary but not incorrect to include. All correct approaches will achieve full marks.

BlackTom AI · Accepted Answer

dp[i] = 1 + max{ dp[k] : 1 ≤ k < i and w[k] > w[i] }, with the convention that max of an empty set is 0. Equivalently (case form that avoids overlap): 1) If there exists k < i with w[k] > w[i], then dp[i] = 1 + max_{1 ≤ k < i, w[k] > w[i]} dp[k]. 2) If no such k exists, then dp[i] = 1. 1. Relevant knowledge points and concepts - This is a longest strictly decreasing subsequence (by weight) problem on a sequence with order constraints; dp[i] is defined to include box i. - Standard DP: each dp[i] equals 1 (the box itself) plus the best chain that can come before it from earlier, heavier boxes. - Max over empty set = 0 makes the unified form work: dp[i] = 1 + max(... ∪ {0}). 2. Complete reasoning process step by step - Any valid stack that ends at i consists of box i on top of a valid stack that ends at some earlier box k with w[k] > w[i]. - Among all such k, choose the one with largest dp[k] to maximize the total count. - If no earlier heavier box exists, the best you can do that includes i is just i alone, so dp[i] = 1. 3. Intermediate steps / calculation demonstration - For each i: collect S = { dp[k] | 1 ≤ k < i and w[k] > w[i] }. - If S ≠ ∅: dp[i] = 1 + max(S). - If S = ∅: dp[i] = 1 (equivalently 1 + max(S ∪ {0}) = 1 + 0 = 1). 4. Why the correct answer is correct - The expression captures exactly the requirement “must include box i,” enforces strict decreasing (w[k] > w[i]), respects order (k < i), and chooses the longest preceding chain via the max. The base case is handled cleanly by the empty-set convention. 5. Why other options would be incorrect (briefly) - Using ≥ instead of > would allow equal weights, violating “strictly decreasing.” - Omitting the +1 would not count box i itself. - Allowing k ≥ i or k = i would violate the order or reuse the same box. - Not handling the empty predecessor case would leave dp[i] undefined when no heavier earlier boxes exist. 6. Conclusion The recurrence with the two non-overlapping cases above is correct; therefore the correct options are c. and j.

类似问题

Following algorithms is a perfect example on how we implemented dynamic programming to generate FibonacciNumber. FibonacciNumber(termIndex) { if (termIndex == 0) return 0 else if (termIndex == 1) return 1 else return FibonacciNumber(termIndex - 1) + FibonacciNumber(termIndex - 2) }

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单