Overview

In this problem, we are given an \(H \times W\) grid and asked to determine whether a robot that alternates between vertical and horizontal moves can reach \((H, W)\) from \((1, 1)\), and if so, find the minimum number of moves required. We can solve this using Breadth-First Search (BFS) on an extended grid, adding the information of “whether the next move must be vertical or horizontal” to the state.

Analysis

A Naive First Thought

In a standard “shortest path on a grid with obstacles” problem, we can solve it in \(O(HW)\) by running a BFS where each cell \((i, j)\) is a vertex. However, this problem introduces the constraint that moves must “alternate between vertical and horizontal.”

If we simply use \((i, j)\) as the state in our BFS, it will lead to WA (Wrong Answer). This is because even when arriving at the exact same cell \((i, j)\), there are \(2\) distinct situations:

• “The previous move was vertical, so the next move must be horizontal”
• “The previous move was horizontal, so the next move must be vertical”

The moves that can be made from this cell differ completely depending on which situation applies. If we mark \((i, j)\) as visited based on the first path that reaches it, we would prune any path that reaches \((i, j)\) in the other state, potentially missing a valid path to the goal or yielding a suboptimal answer.

Concrete Example

0 0
1 0

Here \(H=W=2\). From \((1,1)\), we make a horizontal move to \((1,2)\) (1st move). Next, we must move vertically, so we make a vertical move to \((2,2)\) (2nd move). The answer is \(2\).

On the other hand, for:

0 1
0 0

the path is \((1,1) \to (2,1)\) (vertical) \(\to (2,2)\) (horizontal), which also takes \(2\) moves.

However, consider an example like:

0 0 0
0 1 0
0 0 0

Because the center cannot be traversed, depending on the route taken, one might want to make two consecutive vertical moves. Here, the same cell could potentially be reached in both states (“next is vertical” and “next is horizontal”), which makes it necessary to distinguish and manage these states separately.

Solution: Expanding the State Space

Thus, we define the vertices (states) as:

\[(i,\ j,\ t) \quad (t \in \{0, 1\})\]

where: - \(t = 0\): The next move must be vertical (= the previous move was horizontal, or it is the very first move) - \(t = 1\): The next move must be horizontal (= the previous move was vertical, or it is the very first move)

The transitions are as follows:

• \((i,j,0) \to (i\pm1,\ j,\ 1)\) (moving vertically means the next move must be horizontal)
• \((i,j,1) \to (i,\ j\pm1,\ 0)\) (moving horizontally means the next move must be vertical)

Naturally, the destination cell must be within the grid boundaries and satisfy \(A = 0\) (an empty cell).

Since the first move can be in either direction, we initialize both \((0,0,0)\) and \((0,0,1)\) as starting states with distance \(0\). The goal is reached if we arrive at either \((H-1,W-1,0)\) or \((H-1,W-1,1)\), so we take the minimum of the distances to these two states (using 0-indexed coordinates).

Since all edges have a weight of \(1\), running BFS on this extended graph finds the shortest path.

Algorithm

1. Set up a state space \((i, j, t)\) with \(2HW\) vertices and initialize the dist array with \(\infty\) (unvisited).
2. Set dist[(0,0,0)] = dist[(0,0,1)] = 0 and push both into the queue.
3. Pop a state \((i,j,t)\) from the queue:
   • If \(t = 0\), attempt transitions in the \(2\) vertical directions (up and down); if \(t = 1\), attempt transitions in the \(2\) horizontal directions (left and right).
   • If the destination is within bounds, is an empty cell, and has not yet been visited, record its distance as \(+1\) from the current state and push it into the queue.
4. After the BFS terminates, the answer is min(dist[(H-1,W-1,0)], dist[(H-1,W-1,1)]). If both are still \(\infty\), output NONE.

Because BFS has the property that “states visited earlier always have a shorter (or equal) distance,” once a distance is fixed, it will never be updated again. As a result, each state is processed at most once.

Note that due to the relationship between the parity of \((i+j)\) and \(t\), only about half of the \(2HW\) states are actually reachable. However, without worrying about this detail, simply running BFS over the \(2\)-fold state space is more than fast enough.

Complexity

• Time Complexity: \(O(HW)\) — There are \(2HW\) states, and each state has at most \(2\) outgoing edges, yielding \(O(HW)\). With \(H, W \le 1000\), there are at most about \(2 \times 10^6\) states.
• Space Complexity: \(O(HW)\) — \(O(HW)\) for the dist array and the grid.

Implementation Notes

• State Encoding: Flattening \((i,j,t)\) into a single 1D integer as ((i*W + j)*2 + t) allows you to track distances with a single vector<int>, which is very fast. When decoding, you can recover the components via t = cur & 1, cell = cur >> 1, i = cell / W, and j = cell % W.

• Don’t Forget to Initialize Both Starts: The first move can be vertical or horizontal, so push both \(t=0\) and \(t=1\) into the queue. Pushing only one will lead to a WA.

• Checking the Goal: Once you reach the goal cell, it does not matter which direction the next move would have been, so take the minimum distance between \(t=0\) and \(t=1\).

• Handling Unreachable Cases: If the distance remains INF, output NONE. Since \(H, W \le 1000\) corresponds to up to \(10^6\) cells, using fast I/O or scanf is recommended (in Python, read everything at once using sys.stdin.read).

• Since this is an unweighted shortest path problem, use a queue (or deque); Dijkstra’s algorithm is unnecessary because all edge weights are \(1\).

  Source Code

#include <bits/stdc++.h>
using namespace std;

int main(){
    int H, W;
    if(scanf("%d %d", &H, &W)!=2) return 0;
    vector<vector<int>> A(H, vector<int>(W));
    for(int i=0;i<H;i++)for(int j=0;j<W;j++) scanf("%d", &A[i][j]);

    // state: (i,j,t) t=0 -> next move must be vertical, t=1 -> next move must be horizontal
    const int INF = INT_MAX;
    vector<int> dist((size_t)H*W*2, INF);
    auto idx=[&](int i,int j,int t){ return ((size_t)i*W + j)*2 + t; };

    deque<int> q;
    // start: at (0,0), first move can be either direction
    dist[idx(0,0,0)] = 0;
    dist[idx(0,0,1)] = 0;
    q.push_back((int)idx(0,0,0));
    q.push_back((int)idx(0,0,1));

    while(!q.empty()){
        int cur = q.front(); q.pop_front();
        int t = cur & 1;
        int cell = cur >> 1;
        int i = cell / W, j = cell % W;
        int d = dist[cur];
        if(t==0){
            // vertical move
            for(int di : {-1, 1}){
                int ni = i + di, nj = j;
                if(ni<0||ni>=H) continue;
                if(A[ni][nj]) continue;
                size_t ns = idx(ni,nj,1);
                if(dist[ns]==INF){ dist[ns]=d+1; q.push_back((int)ns); }
            }
        } else {
            for(int dj : {-1, 1}){
                int ni = i, nj = j + dj;
                if(nj<0||nj>=W) continue;
                if(A[ni][nj]) continue;
                size_t ns = idx(ni,nj,0);
                if(dist[ns]==INF){ dist[ns]=d+1; q.push_back((int)ns); }
            }
        }
    }

    int ans = INF;
    ans = min(dist[idx(H-1,W-1,0)], dist[idx(H-1,W-1,1)]);
    if(ans==INF) printf("NONE\n");
    else printf("%d\n", ans);
    return 0;
}

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This editorial was generated by claude5opus-high.