Overview

This problem asks you to move a robot through a grid-shaped warehouse according to commands, collect items on visited cells, and determine the number of items remaining after all commands are executed.

Analysis

Problem Breakdown

In this problem, the robot operates as follows: 1. Start at cell \((1, 1)\) (or \((0, 0)\) in 0-indexed) 2. Move according to each command (movements outside the grid are ignored) 3. If there is an item at the destination cell, collect it

Is a Naive Approach Sufficient?

This problem can be solved with a naive approach that simulates each command one by one.

• The number of commands \(N\) is at most \(2 \times 10^5\)
• Processing each command takes \(O(1)\) (only position update and item check)

Therefore, the total processing time is \(O(N)\), which is sufficiently fast.

Key Observations

1. Items disappear once collected: Even if the same cell is visited multiple times, nothing happens from the second visit onward
2. We need to find “remaining items”: Simply subtract the number of collected items from the initial total number of items
3. Movements outside the grid are ignored: Don’t forget to perform boundary checks

Algorithm

1. Input Processing: Read the warehouse grid and count the total number of items
2. Initial Position Processing: Place the robot at cell \((0, 0)\) and collect the item if one exists
3. Command Execution: For each command, do the following:
   • Calculate the destination coordinates based on the command
   • If the destination is within the grid, move there and collect the item if one exists (update the grid)
   • If the destination is outside the grid, do nothing
4. Output the Result: Output (total items) − (collected items)

Concrete Example

For example, consider the following state on a 3×3 grid:

#.#
...
#.#

If the commands are RRDD: - Start: \((0,0)\) → collect item (3 remaining) - R: \((0,1)\) → no item - R: \((0,2)\) → collect item (2 remaining) - D: \((1,2)\) → no item - D: \((2,2)\) → collect item (1 remaining)

The number of items remaining at the end is 1.

Complexity

• Time Complexity: \(O(H \times W + N)\)
  • Reading the grid takes \(O(H \times W)\)
  • Executing commands takes \(O(N)\) (each command is \(O(1)\))
• Space Complexity: \(O(H \times W + N)\)
  • Storing the grid takes \(O(H \times W)\)
  • Storing the command string takes \(O(N)\)

Implementation Notes

1. Difference between 0-indexed and 1-indexed: The problem statement uses 1-indexed, but using 0-indexed in the implementation makes array access easier

2. Boundary Check: Before moving, verify that the destination satisfies 0 <= nr < H and 0 <= nc < W

3. Item Collection Processing: Update collected cells to '.' to prevent double-counting on the same cell

4. Processing When Not Moving: Even when a command is ignored, “collect item if present” processing could occur, but since the current cell has already been visited, effectively nothing happens (in this implementation, the collection check is only performed when a move actually occurs)

   Source Code

def solve():
    H, W, N = map(int, input().split())
    
    # 倉庫の状態を読み込む
    grid = []
    total_cargo = 0
    for i in range(H):
        row = input().strip()
        grid.append(list(row))
        total_cargo += row.count('#')
    
    # 命令を読み込む
    T = input().strip()
    
    # ロボットの初期位置
    r, c = 0, 0
    
    # 回収した荷物の数
    collected = 0
    
    # 初期位置の荷物を回収
    if grid[r][c] == '#':
        grid[r][c] = '.'
        collected += 1
    
    # 命令を順番に実行
    for cmd in T:
        nr, nc = r, c
        if cmd == 'U':
            nr = r - 1
        elif cmd == 'D':
            nr = r + 1
        elif cmd == 'L':
            nc = c - 1
        elif cmd == 'R':
            nc = c + 1
        
        # グリッド内かチェック
        if 0 <= nr < H and 0 <= nc < W:
            r, c = nr, nc
            # 荷物があれば回収
            if grid[r][c] == '#':
                grid[r][c] = '.'
                collected += 1
    
    # 残っている荷物の数
    print(total_cargo - collected)

solve()

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