feat: add solution to lc problem: No.3562

solution/3500-3599/3562.Maximum Profit from Trading Stocks with Discounts/README.md

@@ -143,7 +143,39 @@ tags:
 
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-### 方法一
+### 方法一：树形动态规划
+
+对每个节点 $u$，我们维护一个二维数组 $f_u[j][pre]$，表示在以 $u$ 为根的子树中，预算不超过 $j$ 且 $u$ 的上司是否购买了股票（其中 $pre=1$ 表示购买，而 $pre=0$ 表示未购买）的情况下，可以获得的最大利润。那么答案就是 $f_1[\text{budget}][0]$。
+
+对节点 $u$，函数 $\text{dfs}(u)$ 返回一个 $(\text{budget}+1) \times 2$ 的二维数组 $f$，表示在以 $u$ 为根的子树中，不超过预算 $j$ 且 $u$ 的上司是否购买了股票的情况下，可以获得的最大利润。
+
+对 $u$，我们要考虑两件事：
+
+1. 节点 $u$ 本身是否买股票（会占用一部分预算 $\text{cost}$，其中 $\text{cost} = \lfloor \text{present}[u] / (pre + 1) \rfloor$）。并增加利润 $\text{future}[u] - \text{cost}$。
+2. 节点 $u$ 的子节点 $v$ 如何分配预算以最大化利润。我们把每个子节点的 $\text{dfs}(v)$ 看成“物品”，用背包把子树的利润合并到当前 $u$ 的 $\text{nxt}$ 数组中。
+
+具体实现时，我们先初始化一个 $(\text{budget}+1) \times 2$ 的二维数组 $\text{nxt}$，表示当前已经合并了子节点的利润。然后对于每个子节点 $v$，我们递归调用 $\text{dfs}(v)$ 得到子节点的利润数组 $\text{fv}$，并用背包把 $\text{fv}$ 合并到 $\text{nxt}$ 中。
+
+合并公式为：
+
+$$
+\text{nxt}[j][pre] = \max(\text{nxt}[j][pre], \text{nxt}[j - j_v][pre] + \text{fv}[j_v][pre])
+$$
+
+其中 $j_v$ 表示分配给子节点 $v$ 的预算。
+
+合并完所有子节点后的 $\text{nxt}[j][pre]$ 表示在 $u$ 本身尚未决定是否购买股票的情况下，且 $u$ 的上次购买状态为 $pre$ 时，把预算 $j$ 全部用于子节点所能获得的最大利润。
+
+最后，我们决定 $u$ 是否购买股票。
+
+- 如果 $j \lt \text{cost}$，则 $u$ 无法购买股票，此时 $f[j][pre] = \text{nxt}[j][0]$。
+- 如果 $j \geq \text{cost}$，则 $u$ 可以选择购买或不购买股票，此时 $f[j][pre] = \max(\text{nxt}[j][0], \text{nxt}[j - \text{cost}][1] + (\text{future}[u] - \text{cost}))$。
+
+最后返回 $f$ 即可。
+
+答案为 $\text{dfs}(1)[\text{budget}][0]$。
+
+时间复杂度 $O(n \times \text{budget}^2)$，空间复杂度 $O(n \times \text{budget})$。
 
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solution/3500-3599/3562.Maximum Profit from Trading Stocks with Discounts/README_EN.md

@@ -140,7 +140,39 @@ tags:
 
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-### Solution 1
+### Solution 1: Tree Dynamic Programming
+
+For each node $u$, we maintain a 2D array $f_u[j][pre]$, representing the maximum profit that can be obtained in the subtree rooted at $u$ with a budget not exceeding $j$ and whether $u$'s manager purchased stocks (where $pre=1$ means purchased, and $pre=0$ means not purchased). The answer is $f_1[\text{budget}][0]$.
+
+For node $u$, the function $\text{dfs}(u)$ returns a $(\text{budget}+1) \times 2$ 2D array $f$, representing the maximum profit that can be obtained in the subtree rooted at $u$ with a budget not exceeding $j$ and whether $u$'s manager purchased stocks.
+
+For $u$, we need to consider two things:
+
+1. Whether node $u$ itself buys stocks (which will consume part of the budget $\text{cost}$, where $\text{cost} = \lfloor \text{present}[u] / (pre + 1) \rfloor$), and increase the profit by $\text{future}[u] - \text{cost}$.
+2. How to allocate the budget among node $u$'s child nodes $v$ to maximize profit. We treat each child node's $\text{dfs}(v)$ as an "item" and use a knapsack approach to merge the subtree profits into the current $u$'s $\text{nxt}$ array.
+
+In the specific implementation, we first initialize a $(\text{budget}+1) \times 2$ 2D array $\text{nxt}$, representing the profits that have been merged from child nodes. Then for each child node $v$, we recursively call $\text{dfs}(v)$ to get the profit array $\text{fv}$ of the child node, and use a knapsack approach to merge $\text{fv}$ into $\text{nxt}$.
+
+The merge formula is:
+
+$$
+\text{nxt}[j][pre] = \max(\text{nxt}[j][pre], \text{nxt}[j - j_v][pre] + \text{fv}[j_v][pre])
+$$
+
+where $j_v$ represents the budget allocated to child node $v$.
+
+After merging all child nodes, $\text{nxt}[j][pre]$ represents the maximum profit that can be obtained by allocating the entire budget $j$ to child nodes when $u$ itself has not yet decided whether to buy stocks, and $u$'s manager's purchase state is $pre$.
+
+Finally, we decide whether $u$ purchases stocks.
+
+- If $j \lt \text{cost}$, then $u$ cannot purchase stocks, in which case $f[j][pre] = \text{nxt}[j][0]$.
+- If $j \geq \text{cost}$, then $u$ can choose to purchase or not purchase stocks, in which case $f[j][pre] = \max(\text{nxt}[j][0], \text{nxt}[j - \text{cost}][1] + (\text{future}[u] - \text{cost}))$.
+
+Finally, return $f$.
+
+The answer is $\text{dfs}(1)[\text{budget}][0]$.
+
+The time complexity is $O(n \times \text{budget}^2)$, and the space complexity is $O(n \times \text{budget})$.
 
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