Overview

We maintain the “next movement direction” in addition to the “current cell” as part of the state, and find the minimum number of moves using Breadth-First Search (BFS). Since the initial move can be either vertical or horizontal, we prepare both states for the starting cell.

Analysis

Cell Positions Alone Cannot Distinguish States

In standard shortest-path maze problems, a BFS that marks each cell as visited can be used. However, in this problem, even at the same cell, the next allowed movement directions may differ.

For example, if we arrive at a cell via a vertical move, the next move can only be horizontal (left or right). Conversely, if we arrive via a horizontal move, the next move can only be vertical (up or down).

Therefore, if we mark cells as visited solely based on their coordinates, paths with a different next movement direction might be incorrectly pruned from the search, leading to an incorrect answer.

Adding “Next Movement Direction” to the State

Thus, we define the state as follows:

• \((i,j,0)\): Currently at cell \((i,j)\), next move is vertical
• \((i,j,1)\): Currently at cell \((i,j)\), next move is horizontal

Since a vertical move is followed by a horizontal move, and a horizontal move is followed by a vertical move, the last value of the state alternates with each move.

There are \(2\) states per cell, so the total number of states is at most \(2HW\).

If we were to enumerate all possible paths, the search could continue indefinitely because revisiting the same cell is allowed. However, returning to the same state of “(cell, next direction)” allows the exact same set of future moves. Therefore, it is sufficient to process each state only once, upon reaching it in the minimum number of moves.

Using BFS Because All Moves Have Equal Cost

Every move increases the number of moves by \(1\). In such a graph, BFS can explore states in ascending order of the number of moves.

Since the direction of the first move is arbitrary, we initialize the search with the two states at the starting cell:

• \((1,1,0)\)
• \((1,1,1)\)

both at a distance of \(0\).

Algorithm

1. For each empty cell, prepare the two states: “next is vertical” and “next is horizontal”.
2. Mark the two states at the starting cell as visited and push them into the queue. Set the current number of moves to \(0\).
3. Pop a state from the queue and perform the following:
   • If the current position is the goal, output the current number of moves and terminate.
   • If the next direction is vertical, examine the cells above and below:
     • If the destination is an empty cell and the “next is horizontal” state is unvisited, mark it as visited and push it into the queue.
   • If the next direction is horizontal, examine the cells to the left and right:
     • If the destination is an empty cell and the “next is vertical” state is unvisited, mark it as visited and push it into the queue.
4. Once all states with the current number of moves have been processed, increment the number of moves by \(1\).
5. If the queue becomes empty without reaching the goal, output NONE.

Correctness

In the state transitions, moves are only made in the specified direction, after which the next direction is toggled. Thus, any path found during the search is guaranteed to satisfy the alternating vertical/horizontal movement condition.

Conversely, all valid moves that satisfy the condition can be represented by these state transitions. Moreover, because we start the search from both initial states at the origin, paths starting with either a vertical move or a horizontal move are fully explored.

Since BFS pops states in ascending order of the number of moves, the number of moves when a goal state is first popped is guaranteed to be minimal. If the search finishes without finding the goal, no reachable path exists.

Complexity

• Time Complexity: \(O(HW)\)
• Space Complexity: \(O(HW)\)

There are at most \(2HW\) states, and from each state we examine at most \(2\) destination cells. Since each state is pushed to the queue at most once, the total time complexity is \(O(HW)\).

Implementation Details

Adding Sentinels Around the Grid

In the code, the grid is surrounded by impassable cells (sentinels), using an array of size \((H+2)\times(W+2)\).

Letting the width be stride = W + 2, the 1D array index of cell \((i,j)\) is:

\(pos=i\cdot \mathrm{stride}+j\)

The positions of adjacent cells (up, down, left, right) can be calculated as follows:

• Up: pos - stride
• Down: pos + stride
• Left: pos - 1
• Right: pos + 1

Because the outer sentinels are always impassable, there is no need to perform boundary checks for each move.

Combining Passability and Visited Information

In the code, vertical and horizontal are bytearrays with the following meanings:

• vertical[pos] == 1: The cell is empty, and the “next is vertical” state is unvisited
• horizontal[pos] == 1: The cell is empty, and the “next is horizontal” state is unvisited

Obstacles, sentinels, and visited states are all represented by 0. Setting a state to 0 at the moment it is pushed to the queue prevents duplicate additions of the same state.

Representing a State as a Single Integer

Using pos for the position and d for the next direction, a state is represented as:

(pos << 1) | d

Here, d = 0 represents the vertical direction, and d = 1 represents the horizontal direction.

To extract them, state >> 1 gives the position, and state & 1 gives the direction.

Processing the BFS Level by Level (by Number of Moves)

By using for _ in range(len(queue)), only the states currently in the queue at that step are processed. States added during processing are deferred to the next round, so the number of moves can be managed simply by incrementing distance by \(1\) after each round.

Source Code

import sys
from collections import deque


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

    stride = W + 2
    size = (H + 2) * stride
    vertical = bytearray(size)

    for i in range(1, H + 1):
        start = i * stride + 1
        vertical[start:start + W] = bytearray(
            x == b'0' for x in input().split()
        )

    horizontal = vertical[:]
    start = stride + 1
    goal = H * stride + W

    vertical[start] = 0
    horizontal[start] = 0

    queue = deque((start << 1, (start << 1) | 1))
    append = queue.append
    popleft = queue.popleft
    distance = 0

    while queue:
        for _ in range(len(queue)):
            state = popleft()
            pos = state >> 1

            if pos == goal:
                print(distance)
                return

            if state & 1:
                nxt = pos - 1
                if vertical[nxt]:
                    vertical[nxt] = 0
                    append(nxt << 1)

                nxt = pos + 1
                if vertical[nxt]:
                    vertical[nxt] = 0
                    append(nxt << 1)
            else:
                nxt = pos - stride
                if horizontal[nxt]:
                    horizontal[nxt] = 0
                    append((nxt << 1) | 1)

                nxt = pos + stride
                if horizontal[nxt]:
                    horizontal[nxt] = 0
                    append((nxt << 1) | 1)

        distance += 1

    print("NONE")


if __name__ == "__main__":
    main()

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This editorial was generated by gpt-6-astra-medium.