二维数组交替A/B规则最短路径求解：Dijkstra算法是否适用？

Absolutely, Dijkstra's algorithm is completely feasible for this problem—you just need to adjust it to track the "alternation state" alongside each node. Let me break this down for you:

Why Your Original Approach Might Have Hit a Bottleneck #

Chances are you were using a standard 2D distance array (dist[i][j]) that only tracks the shortest path to (i,j) without considering whether you arrived at (i,j) via an A or a B. Since the path requires strict A→B→A→B alternation, two different paths to the same node can be valid (one ending with A, one ending with B) and lead to different optimal paths forward. Ignoring this state is probably what caused your bottleneck.

How to Adapt Dijkstra's Algorithm #

The fix is to model each "state" as a combination of:

• Current coordinates (i,j)
• The type of node you need to move to next (or equivalently, the type of the last node you visited)

Here's a step-by-step breakdown:

1. 3D Distance Array: Use a 3D array dist[i][j][state] where:

   • state = 0: You just arrived at (i,j) via an A node (next step needs to be a B)
   • state = 1: You just arrived at (i,j) via a B node (next step needs to be an A)
     Initialize all values to infinity, then set the starting state based on the top-left node's type. For example, if grid[0][0] is A, set dist[0][0][0] = grid[0][0] (since you've visited an A, next needs to be B).

2. Priority Queue: Your priority queue should store entries with three pieces of info: (totalPathLength, i, j, currentState). This ensures you always process the shortest valid path first, just like standard Dijkstra's.

3. Relaxation Step: For each state you process, check all 4-directional neighbors. If the neighbor's type matches what you need next (e.g., current state is 0, neighbor is B), calculate the new path length. If this new length is shorter than the stored distance for the neighbor's new state, update it and add the new state to the queue.

Example Walkthrough (Matching Your Expected Output of 13) #

Suppose your grid has a valid alternating path that sums to 13—say a sequence like A→B→A→B→A where each node's value adds up to the total. The adapted Dijkstra's will track each state transition, ensuring you never break the alternation rule, and lock in this optimal path efficiently.

Quick Java Code Snippet (Simplified) #

// Define a state class for the priority queue
class NodeState {
    int cost;
    int i;
    int j;
    int nextType; // 0 = need B next, 1 = need A next

    NodeState(int cost, int i, int j, int nextType) {
        this.cost = cost;
        this.i = i;
        this.j = j;
        this.nextType = nextType;
    }
}

// Main algorithm setup
public int shortestAlternatingPath(int[][] grid) {
    int rows = grid.length;
    int cols = grid[0].length;
    int[][][] dist = new int[rows][cols][2];
    // Initialize all distances to infinity
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            Arrays.fill(dist[i][j], Integer.MAX_VALUE);
        }
    }

    PriorityQueue<NodeState> pq = new PriorityQueue<>(Comparator.comparingInt(n -> n.cost));
    // Assume top-left is A (adjust logic if it's B)
    int startVal = grid[0][0];
    dist[0][0][0] = startVal; // Next step needs B
    pq.add(new NodeState(startVal, 0, 0, 0));

    int[][] dirs = {{-1,0}, {1,0}, {0,-1}, {0,1}};
    while (!pq.isEmpty()) {
        NodeState curr = pq.poll();
        // If we've reached the bottom-right, return the cost
        if (curr.i == rows-1 && curr.j == cols-1) {
            return curr.cost;
        }
        // Skip if we've already found a better path to this state
        if (curr.cost > dist[curr.i][curr.j][curr.nextType]) {
            continue;
        }
        // Check all neighbors
        for (int[] dir : dirs) {
            int ni = curr.i + dir[0];
            int nj = curr.j + dir[1];
            if (ni >=0 && ni < rows && nj >=0 && nj < cols) {
                // Check if neighbor matches the required next type
                boolean isValid = (curr.nextType == 0 && grid[ni][nj] == 'B') || 
                                  (curr.nextType == 1 && grid[ni][nj] == 'A');
                if (isValid) {
                    int newCost = curr.cost + grid[ni][nj];
                    int newState = curr.nextType ^ 1; // Flip state (0→1, 1→0)
                    if (newCost < dist[ni][nj][newState]) {
                        dist[ni][nj][newState] = newCost;
                        pq.add(new NodeState(newCost, ni, nj, newState));
                    }
                }
            }
        }
    }
    // If no valid path exists
    return -1;
}

This adapted approach will efficiently find the shortest alternating path just like standard Dijkstra's, but with the extra state tracking to enforce the A→B→A→B rule.

内容的提问来源于stack exchange，提问作者David