<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>

from collections import deque
class Solution:
    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        """
        """
        self.n = len(positions)
        self.directions = [(-2, -1), (-2, 1), (2, -1), (2, 1),
                           (-1, -2), (-1, 2), (1, -2), (1, 2)]
        self.size = 50

        self.dists = [[0] * (self.n+1) for i in range(self.n+1)]
        for i in range(1, self.n+1):
            p_i = positions[i-1]
            self.dists[0][i] = self.calc_min_step(kx, ky, p_i[0], p_i[1])
            self.dists[i][0] = self.dists[0][i]
            for j in range(i+1, self.n+1):
                p_j = positions[j-1]
                self.dists[i][j] = self.calc_min_step(p_i[0], p_i[1],  p_j[0], p_j[1])
                self.dists[j][i] = self.dists[i][j]

        # print(self.dists)
        
        res = 0
        for i in range(1, self.n+1):
            mask = 1 << (i-1)
            res = max(res, self.dists[0][i] + self.min_dfs(i-1, mask))
        return res


    @lru_cache(maxsize = None)
    def calc_min_step(self, x, y, x_e, y_e):
        self.visited_cell = set()
        if x == x_e and y_e == y:
            return 0
        queue = deque([(0, x, y)])
        self.visited_cell.add((x, y))
        while queue:
            d, x, y = queue.popleft()
            for dx, dy in self.directions:
                x_new, y_new = x+dx, y+dy
                if 0 <= x_new < self.size and 0 <= y_new < self.size and (x_new, y_new) not in self.visited_cell:
                    if x_new == x_e and y_e == y_new:
                        return d+1
                    queue.append((d+1, x_new, y_new))
                    self.visited_cell.add((x_new, y_new))
        return 1000000

    @lru_cache(maxsize = None)
    def max_dfs(self, i, mask):
        if mask == 2**self.n-1:
            return 0
        res = 0
        for j in range(self.n):
            if 1 << j & mask == 0:
                new_mask = 1 << j | mask
                # print(self.dists[i+1][j+1], self.min_dfs(j, new_mask))
                res = max(res, self.dists[i+1][j+1] + self.min_dfs(j, new_mask))
        return res

    @lru_cache(maxsize = None)
    def min_dfs(self, i, mask):
        if mask == 2**self.n-1:
            return 0
        res = float("inf")
        for j in range(self.n):
            if 1 << j & mask == 0:
                new_mask = 1 << j | mask
                res = min(res, self.dists[i+1][j+1] + self.max_dfs(j, new_mask))
        return res
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