You are given an m x n grid classroom where a student volunteer is tasked with cleaning up litter scattered around the room. Each cell in the grid is one of the following:
'S': Starting position of the student'L': Litter that must be collected (once collected, the cell becomes empty)'R': Reset area that restores the student's energy to full capacity, regardless of their current energy level (can be used multiple times)'X': Obstacle the student cannot pass through'.': Empty spaceYou are also given an integer energy, representing the student's maximum energy capacity. The student starts with this energy from the starting position 'S'.
Each move to an adjacent cell (up, down, left, or right) costs 1 unit of energy. If the energy reaches 0, the student can only continue if they are on a reset area 'R', which resets the energy to its maximum capacity energy.
Return the minimum number of moves required to collect all litter items, or -1 if it's impossible.
\n
Example 1:
\n\nInput: classroom = ["S.", "XL"], energy = 2
\n\nOutput: 2
\n\nExplanation:
\n\n(0, 0) with 2 units of energy.(1, 0) contains an obstacle 'X', the student cannot move directly downward.(0, 0) → (0, 1) with 1 unit of energy and 1 unit remaining.(0, 1) → (1, 1) to collect the litter 'L'.Example 2:
\n\nInput: classroom = ["LS", "RL"], energy = 4
\n\nOutput: 3
\n\nExplanation:
\n\n(0, 1) with 4 units of energy.(0, 1) → (0, 0) to collect the first litter 'L' with 1 unit of energy used and 3 units remaining.(0, 0) → (1, 0) to 'R' to reset and restore energy back to 4.(1, 0) → (1, 1) to collect the second litter 'L'.Example 3:
\n\nInput: classroom = ["L.S", "RXL"], energy = 3
\n\nOutput: -1
\n\nExplanation:
\n\nNo valid path collects all 'L'.
\n
Constraints:
\n\n1 <= m == classroom.length <= 201 <= n == classroom[i].length <= 20classroom[i][j] is one of 'S', 'L', 'R', 'X', or '.'1 <= energy <= 50'S' in the grid.'L' cells in the grid.给你一个 m x n 的网格图 classroom，其中一个学生志愿者负责清理散布在教室里的垃圾。网格图中的每个单元格是以下字符之一：
'S' ：学生的起始位置'L' ：必须收集的垃圾（收集后，该单元格变为空白）'R' ：重置区域，可以将学生的能量恢复到最大值，无论学生当前的能量是多少（可以多次使用）'X' ：学生无法通过的障碍物'.' ：空白空间同时给你一个整数 energy，表示学生的最大能量容量。学生从起始位置 'S' 开始，带着 energy 的能量出发。
每次移动到相邻的单元格（上、下、左或右）会消耗 1 单位能量。如果能量为 0，学生此时只有处在 'R' 格子时可以继续移动，此区域会将能量恢复到 最大 能量值 energy。
返回收集所有垃圾所需的 最少 移动次数，如果无法完成，返回 -1。
\n\n
示例 1：
\n\n输入: classroom = [\"S.\", \"XL\"], energy = 2
\n\n输出: 2
\n\n解释:
\n\n(0, 0) 开始，带着 2 单位的能量。(1, 0) 有一个障碍物 'X'，学生无法直接向下移动。(0, 0) → (0, 1)，消耗 1 单位能量，剩余 1 单位。(0, 1) → (1, 1)，收集垃圾 'L'。示例 2：
\n\n输入: classroom = [\"LS\", \"RL\"], energy = 4
\n\n输出: 3
\n\n解释:
\n\n(0, 1) 开始，带着 4 单位的能量。(0, 1) → (0, 0)，收集第一个垃圾 'L'，消耗 1 单位能量，剩余 3 单位。(0, 0) → (1, 0)，到达 'R' 重置区域，恢复能量为 4。(1, 0) → (1, 1)，收集第二个垃圾 'L'。示例 3：
\n\n输入: classroom = [\"L.S\", \"RXL\"], energy = 3
\n\n输出: -1
\n\n解释:
\n\n没有有效路径可以收集所有 'L'。
\n\n
提示：
\n\n1 <= m == classroom.length <= 201 <= n == classroom[i].length <= 20classroom[i][j] 是 'S'、'L'、'R'、'X' 或 '.' 之一1 <= energy <= 50'S'。'L' 单元格。(x, y, mask, e, steps), initializing with (sx, sy, 0, energy, 0), and for each move update e (–1 per step), update mask on 'L', reset e=energy on 'R', and return steps when mask == fullMask.",
"Maintain a 3D array bestEnergy[x][y][mask] storing the maximum e seen for each (x,y,mask) and skip any new state with e <= bestEnergy[x][y][mask] to prune."
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