Leetcode 3562. Maximum Profit from Trading Stocks with Discounts

• [Leetcode 3562. Maximum Profit from Trading Stocks with Discounts](#Leetcode 3562. Maximum Profit from Trading Stocks with Discounts)
  • [1. 解题思路](#1. 解题思路)
  • [2. 代码实现](#2. 代码实现)

• 题目链接：3562. Maximum Profit from Trading Stocks with Discounts

1. 解题思路

这一题没有搞定，思路上整体走偏了，看了一下别人的解答，结合deepseek的回答看了半天，终于是搞明白了里面的道道，也是醉了......

这一题我自己的思路就是一个遍历，分别考察每一个用户买与不买的情况下budget的变化以及对其下属员工price的变动，但是遇到了超时的问题。

然后deepseek给到的解答是视为一个背包问题，即不是考察每个员工买不不买的情况，而是考察给每一个员工及其下属员工分配多少的预算，然后看其最大能获得多大的利润。即，给到的最终的动态规划的数组为dp[uid][status][budget]，其中，uid表示用户id；status表示其父节点的购买状态，即直属上司是否有购买行为；budget表示给该用户及其下属所有员工分配的budget；然后dp[uid][status][budget]整体表示当uid用户的直属上司的购买状态为status时，给他及其下属所有员工分配budget预算时，能够获得的最大的profit。

由此一来，我们就可以使用动态规划进行处理了。

2. 代码实现

给出python代码实现如下：

class Solution:
    def maxProfit(self, n: int, present: List[int], future: List[int], hierarchy: List[List[int]], budget: int) -> int:
        tree = defaultdict(list)
        for u, v in hierarchy:
            tree[u - 1].append(v - 1)

        @lru_cache(None)
        def dp(u):
            '''
            input: u : int -> 0-based user id
            output: 
              - dp0: [int] * (budget+1) -> max profit for each budget if parent node was not bought
              - dp1: [int] * (budget+1) -> max profit for each budget if parent node was bought
            '''
            child_dp = [dp(v) for v in tree[u]]
            dp0 = [0] * (budget+1) # parent not buy
            dp1 = [0] * (budget+1) # parent buy
            for parent_bought, max_profits in [(0, dp0), (1, dp1)]:
                cost = present[u] if parent_bought == 0 else present[u] // 2
                profit = future[u] - cost
                
                dpA = [0] + [-math.inf] * (budget) # u not buy
                dpB = [-math.inf] * (budget+1) # u buy
                if cost <= budget:
                    dpB[cost] = profit
                for case0, case1 in child_dp:
                    new_dpA = [-math.inf] * (budget+1) # u not buy
                    for bgt in range(budget+1):
                        if dpA[bgt] == -math.inf:
                            continue
                        for k in range(budget-bgt+1):
                            if case0[k] == -math.inf:
                                continue
                            new_dpA[bgt+k] = max(new_dpA[bgt+k], dpA[bgt] + case0[k])
                    dpA = new_dpA
                    
                    new_dpB = [-math.inf] * (budget+1) # u buy
                    for bgt in range(budget+1):
                        if dpB[bgt] == -math.inf:
                            continue
                        for k in range(budget-bgt+1):
                            if case1[k] == -math.inf:
                                continue
                            new_dpB[bgt+k] = max(new_dpB[bgt+k], dpB[bgt] + case1[k])
                    dpB = new_dpB
                    
                for bgt in range(budget+1):
                    max_profits[bgt] = max(dpA[bgt], dpB[bgt])
                    
            return dp0, dp1
                    
                
        
        max_profits, _ = dp(0)
        return max(max_profits)

提交代码评测得到：耗时2362ms，占用内存22.8MB。