Overview

This problem asks whether a robot that alternates between vertical and horizontal moves on an \(H \times W\) grid can reach \((H,W)\) from \((1,1)\), and if possible, to find the minimum number of moves required. It can be solved using Breadth-First Search (BFS) on a state graph where each vertex represents a “(cell, last move direction)” pair.

Analysis

Why Naive BFS Fails

In a standard shortest path problem, running BFS with each vertex being a “cell \((i,j)\)” would suffice. However, in this problem, the directions in which the robot can move next depend on the direction of the previous move, meaning that being at the same cell can represent different situations.

For example, when at cell \((2,2)\): - If the previous move was vertical \(\to\) the robot can only move horizontally (left or right) next. - If the previous move was horizontal \(\to\) the robot can only move vertically (up or down) next.

In other words, the cell coordinates alone do not determine how the robot can move next. If we run BFS using only coordinates as the state, we might allow transitions that are not actually valid (WA), or overwrite a cell with an earlier reached state and miss the optimal solution.

Solution: Expanding the State

Therefore, we expand each vertex into a pair:

\[(\text{cell position},\ \text{type of previous move direction})\]

Since there are only \(2\) types of directions—”vertical” and “horizontal”—the number of vertices is \(2HW\).

• State \((\,(i,j),\ \text{vertical}\,)\): The state after a vertical move \(\to\) can only transition horizontally (left/right) next.
• State \((\,(i,j),\ \text{horizontal}\,)\): The state after a horizontal move \(\to\) can only transition vertically (up/down) next.

Since all edges have a weight of \(1\) (\(1\) move = cost \(1\)), the shortest distance can be found using BFS.

Handling the Start and Goal

• Start: Since the first move can be either vertical or horizontal, we push both \((\,(1,1),\text{vertical}\,)\) and \((\,(1,1),\text{horizontal}\,)\) into the BFS queue with distance \(0\) (we interpret starting from the “previous was vertical” state as allowing a horizontal move next, and vice versa).
• Goal: Upon reaching \((H,W)\), the direction of the final move can be either vertical or horizontal. Therefore, the answer is the smaller of the distances to \((\,(H,W),\text{vertical}\,)\) and \((\,(H,W),\text{horizontal}\,)\). If both are unvisited, output NONE.

Note that although the problem allows visiting the same cell multiple times, in the augmented state graph each state only needs to be visited once, so no special handling is required (even for the same cell, different direction types correspond to different vertices, allowing each cell to potentially be visited up to twice).

Algorithm

1. Read the grid as a 1D array g (cell \((i,j)\) has index \((i-1)W + (j-1)\)).
2. Prepare a distance array dist of length \(2HW\), initialized to \(-1\) (unvisited).
   • Index \(u < HW\): at cell \(u\), previous move was vertical (moves horizontally next)
   • Index \(u \ge HW\): at cell \(u - HW\), previous move was horizontal (moves vertically next)
3. Set dist[0] = 0 and dist[HW] = 0, and push these two vertices into the queue.
4. Run BFS:
   • If the state is “move horizontally next”, transition to the adjacent left \((u-1)\) and right \((u+1)\) cells. The resulting state type becomes “horizontal”.
   • If the state is “move vertically next”, transition to the adjacent up \((u-W)\) and down \((u+W)\) cells. The resulting state type becomes “vertical”.
   • If the destination is within bounds, not an obstacle, and unvisited, update its distance and push it to the queue.
5. When the target cell \(HW-1\) is reached in either state, that distance is the answer (since it is BFS, the first arrival is minimal). If it cannot be reached, output NONE.

Complexity

With \(2HW \le 2 \times 10^6\) vertices and an out-degree of \(2\) for each vertex, the number of edges is also \(O(HW)\).

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

Implementation Notes

• 1D Array Flattening: With \(H, W \le 1000\), there are up to \(10^6\) cells and \(2 \times 10^6\) states. Using a 2D list of lists in Python can be slow and memory-intensive, so using a 1D list of length \(2HW\) for dist and a bytearray for the grid is fast and memory-efficient.

• Boundary Checks for Horizontal Moves: When flattened to 1D, the column of index \(u\) is u % W. If u % W == 0, it is at the left edge and cannot move left; if u % W == W-1, it is at the right edge and cannot move right. Forgetting this check causes a bug where the robot wraps around rows (for vertical moves, bounds checks like u - W >= 0 and u + W < N are sufficient).

• Fast I/O: Since there are up to \(10^6\) numbers to read, reading all input at once with sys.stdin.buffer.read().split() and directly comparing with the byte string b'1' (skipping int() conversions) is very fast.

• Queue Implementation: While collections.deque can be used, a list with a head index pointer (q.append and head += 1) is also fast enough and is used here.

• Early Exit: You may terminate the BFS as soon as the target cell is reached for the first time. However, to be safe in case you do not terminate early, taking the minimum of the distances to the two states of \((H,W)\) at the end is also fine.

• Lower Bound of the Answer: The number of moves from \((1,1)\) to \((H,W)\) is at least \((H-1)+(W-1)\), but because the alternating constraint may require extra back-and-forth moves in some cases, it cannot be determined by a simple formula. Always use BFS to compute it.

  Source Code

import sys

def main():
    data = sys.stdin.buffer.read().split()
    H = int(data[0]); W = int(data[1])
    N = H * W
    g = bytearray(N)
    i = 0
    off = 2
    for k in range(N):
        if data[off + k] == b'1':
            g[k] = 1
    dist = [-1] * (2 * N)
    target = N - 1
    dist[0] = 0
    dist[N] = 0
    q = [0, N]
    head = 0
    ans = -1
    while head < len(q):
        u = q[head]; head += 1
        du = dist[u] + 1
        if u < N:
            # last move vertical -> now horizontal
            j = u % W
            if j:
                v = u - 1
                if g[v] == 0:
                    w = N + v
                    if dist[w] < 0:
                        dist[w] = du
                        if v == target:
                            ans = du
                            break
                        q.append(w)
            if j + 1 < W:
                v = u + 1
                if g[v] == 0:
                    w = N + v
                    if dist[w] < 0:
                        dist[w] = du
                        if v == target:
                            ans = du
                            break
                        q.append(w)
        else:
            c = u - N
            # last move horizontal -> now vertical
            if c >= W:
                v = c - W
                if g[v] == 0:
                    if dist[v] < 0:
                        dist[v] = du
                        if v == target:
                            ans = du
                            break
                        q.append(v)
            v = c + W
            if v < N:
                if g[v] == 0:
                    if dist[v] < 0:
                        dist[v] = du
                        if v == target:
                            ans = du
                            break
                        q.append(v)
    if ans < 0:
        d1 = dist[target]
        d2 = dist[N + target]
        if d1 < 0 and d2 < 0:
            sys.stdout.write("NONE\n")
            return
        if d1 < 0:
            ans = d2
        elif d2 < 0:
            ans = d1
        else:
            ans = d1 if d1 < d2 else d2
    sys.stdout.write(str(ans) + "\n")

main()

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