廣度優先搜索解決迷宮問題花費的時間太長

Question

我有一個像這樣的大而開放的迷宮：

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“s”是起點。 並且有多個目的地點“x”。 我只需要找到一個目的地。 如果目的地靠近起點，BFS 算法可以很快找到解決方案。 如果它們像上面的例子一樣離得更遠，則需要無窮無盡的時間。 所以我的問題是：a) 算法對這種特定類型的迷宮是否不利，我應該使用 A* 或類似的東西。 b) 我的實施不好嗎？

執行：

public class BFS {

    public static String getPath(String[][] map) {
        String[] ways = { "L", "R", "U", "D" }; // directions to go
        Queue<String> q = new LinkedList<>();
        q.offer("");
        String path = "";

        while (!foundBait(map, path)) {
            path = q.poll();

            for (String s : ways) {
                String newPath = path + s;
                if (valid(map, newPath))
                    q.offer(newPath);
            }
        }
        return path;
    }

    private static boolean foundBait(String[][] map, String moves) {
        int xStart = 0;
        int yStart = 0;

        for (int y = 0; y < map.length; y++)
            for (int x = 0; x < map[0].length; x++)
                if (map[y][x].equals("s")) {
                    xStart = x;
                    yStart = y;
                }

        int i = xStart;
        int j = yStart;

        for (int s = 0; s < moves.length(); s++) {

            if (moves.charAt(s) == "L".charAt(0))
                i--;
            else if (moves.charAt(s) == "R".charAt(0))
                i++;
            else if (moves.charAt(s) == "U".charAt(0))
                j--;
            else if (moves.charAt(s) == "D".charAt(0))
                j++;

        }

        if (map[j][i].equals("x"))
            return true;

        return false;
    }

    private static boolean valid(String[][] map, String moves) {
        int xStart = 0;
        int yStart = 0;

        for (int y = 0; y < map.length; y++)
            for (int x = 0; x < map[0].length; x++)
                if (map[y][x].equals("s")) {
                    xStart = x;
                    yStart = y;
                }

        int i = xStart;
        int j = yStart;

        for (int s = 0; s < moves.length(); s++) {

            if (moves.charAt(s) == "L".charAt(0))
                i--;
            else if (moves.charAt(s) == "R".charAt(0))
                i++;
            else if (moves.charAt(s) == "U".charAt(0))
                j--;
            else if (moves.charAt(s) == "D".charAt(0))
                j++;

            if (!(0 <= i && i < map[0].length && 0 <= j && j < map.length))
                return false;
            else if (map[j][i].equals("#") || map[j][i].equals("-"))
                return false;

        }
        return true;
    }

}

Answer 1

如評論中所述，問題不在於標記添加到路徑的節點，解決方案是使用第二個矩陣進行標記。