<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 bfs(self, pos: List[int]) -> List[List[int]]:
        res = [[self.inf]*self.n for _ in range(self.n)]
        visited = [[False]*self.n for _ in range(self.n)]
        
        queue = deque()
        queue.append([pos[0], pos[1], 0])
        visited[pos[0]][pos[1]] = True
        res[pos[0]][pos[1]] = 0
        
        while queue:
            i,j,d = queue.popleft()
            for x,y in self.dirs:
                ii,jj = i+x,j+y
                if not (0<=ii<self.n and 0<=jj<self.n):
                    continue
                if visited[ii][jj]:
                    continue
                queue.append([ii,jj,d+1])
                res[ii][jj] = d+1
                visited[ii][jj] = True
        return res
        
    
    def dfs(self, turn: bool, mask: int, x: int, y: int) -> int:
        if x==self.kx and y==self.ky:
            cur_idx = -1
        else:
            cur_idx = self.pos_map[(x,y)]
        if mask == self.terminal:
            return 0
        elif not turn and (mask, cur_idx) in self.max_cache:
            return self.max_cache[(mask, cur_idx)]
        elif turn and (mask, cur_idx) in self.min_cache:
            return self.min_cache[(mask, cur_idx)]
        else:
            # 1. Alice's turn
            if not turn:
                res = -1
                for idx in range(len(self.routes)):
                    if (mask & (1<<idx)) != 0:
                        continue
                    next_mask = (mask | (1<<idx))
                    steps = self.routes[idx][x][y]
                    tmp = steps + self.dfs(not turn, next_mask, self.rev_pos[idx][0], self.rev_pos[idx][1])
                    res = max(res, tmp)
                self.max_cache[(mask, cur_idx)] = res
                return res
            # 2. Bob's turn
            else:
                res = self.inf
                for idx in range(len(self.routes)):
                    if (mask & (1<<idx)) != 0:
                        continue
                    next_mask = (mask | (1<<idx))
                    steps = self.routes[idx][x][y]
                    tmp = steps + self.dfs(not turn, next_mask, self.rev_pos[idx][0], self.rev_pos[idx][1])
                    res = min(res, tmp)
                self.min_cache[(mask, cur_idx)] = res
                return res
    
    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        self.kx = kx
        self.ky = ky
        self.dirs = [[-1,2],[-1,-2],[-2,1],[-2,-1],[1,2],[1,-2],[2,1],[2,-1]]
        self.n = 50
        self.inf = 10**10
        self.terminal = 2**len(positions)-1
        
        self.routes = {}
        self.rev_pos = {}
        self.pos_map = {}
        for idx, position in enumerate(positions):
            res = self.bfs(position)
            self.routes[idx] = res
            self.rev_pos[idx] = position
            self.pos_map[(position[0], position[1])] = idx
        
        self.min_cache = {}
        self.max_cache = {}
        
        res = self.dfs(False, 0, kx, ky)
        
        return res
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