<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:
        pos = [[False] * 50 for _ in range(50)]
        for x, y in positions:
            pos[x][y] = True
        pos[kx][ky] = True

        def min_moves(x, y):
            dp = [[0] * 50 for _ in range(50)]
            dp[x][y] = 0
            queue = deque([(x, y)])
            moves = 0
            visited = [[False] * 50 for _ in range(50)]
            visited[x][y] = True
            directions = ((-2, 1), (-1, 2), (1, 2), (2, 1), (2, -1), (1, -2), (-1, -2), (-2, -1))
            pawns = -1
        
            while queue:
                for _ in range(len(queue)):
                    i, j = queue.popleft() 

                    dp[i][j] = moves
                    if pos[i][j]:
                        pawns += 1
                        if pawns == len(positions):
                            return dp
                
                    for di, dj in directions:
                        ni, nj = i + di, j + dj
                        if ni < 0 or ni >= 50:
                            continue
                        if nj < 0 or nj >= 50:
                            continue
                        if visited[ni][nj]:
                            continue
                        visited[ni][nj] = True
                        queue.append((ni, nj))
            
                moves += 1
        
            return dp
        
        n = len(positions)
        dis = [[0] * (n + 1) for _ in range(n + 1)]
        for i, (x, y) in enumerate(positions):
            dp = min_moves(x, y)
            for j in range(i, n):
                xj, yj = positions[j]
                dis[i][j] = dp[xj][yj]
                dis[j][i] = dp[xj][yj]
            dis[i][-1] = dp[kx][ky]
            dis[-1][i] = dp[kx][ky]
        
        cache = [[[None] * (1 << n) for _ in range(n + 1)] for _ in range(n)]

        # @cache
        def backtrack(turn, prev, state):
            if turn == n:
                return 0
            
            if cache[turn][prev][state] is not None:
                return cache[turn][prev][state]

            max_dis = 0
            min_dis = inf

            for i in range(n):
                if (1 << i) & state:
                    continue
                d = dis[prev][i] + backtrack(turn + 1, i, state | (1 << i))
                if d > max_dis:
                    max_dis = d
                if d < min_dis:
                    min_dis = d
            
            if turn % 2 == 0:
                ans = max_dis
            else:
                ans = min_dis
            
            cache[turn][prev][state] = ans
            return ans
        
        return backtrack(0, -1, 0)
2906