<p>There is a <code>50 x 50</code> chessboard with <strong>one</strong> knight and some pawns on it. You are given two integers <code>kx</code> and <code>ky</code> where <code>(kx, ky)</code> denotes the position of the knight, and a 2D array <code>positions</code> where <code>positions[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> denotes the position of the pawns on the chessboard.</p>

<p>Alice and Bob play a <em>turn-based</em> game, where Alice goes first. In each player&#39;s turn:</p>

<ul>
	<li>The player <em>selects </em>a pawn that still exists on the board and captures it with the knight in the <strong>fewest</strong> possible <strong>moves</strong>. <strong>Note</strong> that the player can select <strong>any</strong> pawn, it <strong>might not</strong> be one that can be captured in the <strong>least</strong> number of moves.</li>
	<li><span>In the process of capturing the <em>selected</em> pawn, the knight <strong>may</strong> pass other pawns <strong>without</strong> capturing them</span>. <strong>Only</strong> the <em>selected</em> pawn can be captured in <em>this</em> turn.</li>
</ul>

<p>Alice is trying to <strong>maximize</strong> the <strong>sum</strong> of the number of moves made by <em>both</em> players until there are no more pawns on the board, whereas Bob tries to <strong>minimize</strong> them.</p>

<p>Return the <strong>maximum</strong> <em>total</em> number of moves made during the game that Alice can achieve, assuming both players play <strong>optimally</strong>.</p>

<p>Note that in one <strong>move, </strong>a chess knight has eight possible positions it can move to, as illustrated below. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.</p>

<p><img src="https://assets.leetcode.com/uploads/2024/08/01/chess_knight.jpg" style="width: 275px; height: 273px;" /></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">kx = 1, ky = 1, positions = [[0,0]]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/16/gif3.gif" style="width: 275px; height: 275px;" /></p>

<p>The knight takes 4 moves to reach the pawn at <code>(0, 0)</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">kx = 0, ky = 2, positions = [[1,1],[2,2],[3,3]]</span></p>

<p><strong>Output:</strong> <span class="example-io">8</span></p>

<p><strong>Explanation:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2024/08/16/gif4.gif" style="width: 320px; height: 320px;" /></strong></p>

<ul>
	<li>Alice picks the pawn at <code>(2, 2)</code> and captures it in two moves: <code>(0, 2) -&gt; (1, 4) -&gt; (2, 2)</code>.</li>
	<li>Bob picks the pawn at <code>(3, 3)</code> and captures it in two moves: <code>(2, 2) -&gt; (4, 1) -&gt; (3, 3)</code>.</li>
	<li>Alice picks the pawn at <code>(1, 1)</code> and captures it in four moves: <code>(3, 3) -&gt; (4, 1) -&gt; (2, 2) -&gt; (0, 3) -&gt; (1, 1)</code>.</li>
</ul>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">kx = 0, ky = 0, positions = [[1,2],[2,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>Alice picks the pawn at <code>(2, 4)</code> and captures it in two moves: <code>(0, 0) -&gt; (1, 2) -&gt; (2, 4)</code>. Note that the pawn at <code>(1, 2)</code> is not captured.</li>
	<li>Bob picks the pawn at <code>(1, 2)</code> and captures it in one move: <code>(2, 4) -&gt; (1, 2)</code>.</li>
</ul>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= kx, ky &lt;= 49</code></li>
	<li><code>1 &lt;= positions.length &lt;= 15</code></li>
	<li><code>positions[i].length == 2</code></li>
	<li><code>0 &lt;= positions[i][0], positions[i][1] &lt;= 49</code></li>
	<li>All <code>positions[i]</code> are unique.</li>
	<li>The input is generated such that <code>positions[i] != [kx, ky]</code> for all <code>0 &lt;= i &lt; positions.length</code>.</li>
</ul>

class Solution:
  def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
    n = 50
    n_positions = len(positions)
    position_indices = [[-1 for _ in range(n)] for _ in range(n)]
    for i in range(n_positions):
      position_indices[positions[i][0]][positions[i][1]] = i
      
    def bfs(start_x: int, start_y: int, distances: List[int]):
      directions = ((2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (1, -2), (-1, 2), (-1, -2))
      n_directions = 8;
      q = deque()
      visited = [[False for _ in range(n)] for _ in range(n)]
      q.append((start_x, start_y))
      visited[start_x][start_y] = True
      distance = -1
      reached = 0
      while len(q) > 0 and reached < n_positions:
        distance += 1
        for _ in range(len(q)):
          x, y = q.popleft()
          if position_indices[x][y] != -1:
            distances[position_indices[x][y]] = distance
            reached += 1
            if reached == n_positions:
              break
          for dx, dy in directions:
            nx = x + dx
            ny = y + dy
            if -1 < nx < n and -1 < ny < n and not visited[nx][ny]:
              q.append((nx, ny))
              visited[nx][ny] = True
    
    distance_knight_to_positions = [0 for _ in range(n_positions)]
    bfs(kx, ky, distance_knight_to_positions)
    distances = [[0 for _ in range(n_positions)] for _ in range(n_positions)]
    for i in range(n_positions):
      x, y = positions[i]
      bfs(x, y, distances[i])
      
    def get_bit_ones(num: int) -> int:
      ret = 0
      while num > 0:
        num &= num - 1
        ret += 1
      return ret
    
    #print(distance_knight_to_positions)
    #print(distances)


    
    layouts = 1 << n_positions
    mask = layouts - 1
    dp = [[0 for _ in range(n_positions)] for _ in range(layouts)]
    for layout in range(1, layouts - 1):
      bit_ones = get_bit_ones(mask ^ layout)
      for i in range(n_positions):
        if (((mask^layout) >> i) & 1) == 1:
          dp[layout][i] = inf if (bit_ones & 1) == 1 else 0
          func = min if (bit_ones & 1) == 1 else max
          d = distances[i]
          for j in range(n_positions):
            if ((layout >> j) & 1) == 1:
              dp[layout][i] = func(dp[layout][i], dp[layout ^ (1 << j)][j] + d[j])
    ret = 0
    for i in range(n_positions):
      ret = max(ret, dp[mask ^ (1 << i)][i] + distance_knight_to_positions[i])
    return ret
    #print(dp)
    #return max(dp[layouts - 1])# if (n_positions & 1) == 1 else max(dp[layouts - 1])


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