Overview

Simulate the robot’s movement exactly as instructed. If there is a package (#) on the cell the robot visits, pick it up on the spot and remove it. Finally, output the number of remaining packages.

Analysis

What we need for this problem is “the number of # that were not picked up in the end.”
There are at most \(N=2\times 10^5\) instructions, but the robot moves at most 1 cell per instruction, so if we can process each instruction in \(O(1)\), the overall solution is fast enough.

The key observations are:

• If the robot is on a cell with #, that package is picked up and no longer exists.
• Even if the same cell is visited multiple times, nothing happens after the first pickup.
  → In other words, we just need to record “already picked up.”

A naive approach like “scanning the entire grid after each instruction to count remaining #” would be \(O(NHW)\) in the worst case and would definitely TLE.
Instead:

• Count the total number of # at the beginning and store it in total
• When the robot arrives at a cell with #, change that cell to . (picked up) and decrement total by 1

This eliminates the need to rescan the entire grid.

Algorithm

1. Store the input grid \(S\) and count the initial number of #, storing it in total.
2. Place the robot at the top-left \((1,1)\) (or \((0,0)\) with 0-indexing).
3. If there is a # at the initial position, pick it up, remove it from the grid, and do total -= 1.
4. Process the instruction sequence \(T\) from the beginning in order.
   • For U/D/L/R, update the coordinates if the destination is within the grid; otherwise ignore (keep coordinates unchanged).
   • After moving (or staying), if the current cell has #, pick it up, remove it, and do total -= 1.
5. After all instructions, output total.

With this method, “already picked up” is represented by directly modifying the grid (#→.), which guarantees we never decrement twice for the same cell.

Complexity

• Time complexity: \(O(HW + N)\) (initial # count is \(HW\), instruction processing is \(N\))
• Space complexity: \(O(HW)\) (storing the grid)

Implementation Notes

• Using 0-indexed coordinates (r,c with \(0 \le r < H,\, 0 \le c < W\)) makes boundary checking straightforward.

• Rather than recounting the remaining number each time, decrementing by 1 from the initial total each time a package is picked up is much faster.

• In the provided code, for optimization, the grid is stored as a bytearray boolean array, and the instruction sequence is also processed as bytes (ord('U')=85, etc.). This allows stable processing of \(N=2\times 10^5\) even in Python.

  Source Code

import sys

def main():
    input = sys.stdin.buffer.readline
    H, W, N = map(int, input().split())

    grid = []
    total = 0
    for _ in range(H):
        s = input().strip()
        total += s.count(b'#')
        row = bytearray(b == 35 for b in s)  # 35 == ord('#')
        grid.append(row)

    T = input().strip()

    r = c = 0
    if grid[0][0]:
        grid[0][0] = 0
        total -= 1

    for b in T:
        if b == 85:  # 'U'
            if r > 0:
                r -= 1
        elif b == 68:  # 'D'
            if r + 1 < H:
                r += 1
        elif b == 76:  # 'L'
            if c > 0:
                c -= 1
        else:  # 'R'
            if c + 1 < W:
                c += 1

        if grid[r][c]:
            grid[r][c] = 0
            total -= 1

    print(total)

if __name__ == "__main__":
    main()

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This editorial was generated by gpt-5.2-high.