<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 typing import List
from collections import deque
from functools import lru_cache

class Solution:
    def bfs_knight(self, start, end):
        """使用 BFS 计算骑士从 `start` 到 `end` 的最小移动步数。"""
        directions = [(2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (1, -2), (-1, 2), (-1, -2)]
        queue = deque([start])
        visited = set([start])
        distance = {start: 0}
        
        while queue:
            x, y = queue.popleft()
            if (x, y) == end:
                return distance[(x, y)]
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                if 0 <= nx < 50 and 0 <= ny < 50 and (nx, ny) not in visited:
                    visited.add((nx, ny))
                    queue.append((nx, ny))
                    distance[(nx, ny)] = distance[(x, y)] + 1
        
        return float('inf')

    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        num_pawns = len(positions)
        knight_pos = (kx, ky)
        
        # 预处理骑士到每个pawn的距离，以及每对pawn之间的距离
        knight_to_pawn_dist = [self.bfs_knight(knight_pos, tuple(pawn)) for pawn in positions]
        pawn_to_pawn_dist = [[float('inf')] * num_pawns for _ in range(num_pawns)]
        
        for i in range(num_pawns):
            for j in range(num_pawns):
                if i != j:
                    pawn_to_pawn_dist[i][j] = self.bfs_knight(tuple(positions[i]), tuple(positions[j]))
        
        @lru_cache(None)
        def dfs(mask, turn, current_pawn):
            """记忆化递归
            mask: 当前剩余的pawns状态（位掩码）
            turn: 当前是Alice的回合(1)还是Bob的回合(0)
            current_pawn: 当前骑士所在的棋子位置（-1表示还未开始）
            """
            if mask == 0:  # 没有剩余的棋子了
                return 0
            
            if turn == 1:  # Alice's turn, maximize the moves
                max_moves = -float('inf')
                for i in range(num_pawns):
                    if mask & (1 << i):  # 如果第i个pawn还在
                        next_mask = mask ^ (1 << i)
                        move_cost = knight_to_pawn_dist[i] if current_pawn == -1 else pawn_to_pawn_dist[current_pawn][i]
                        max_moves = max(max_moves, dfs(next_mask, 0, i) + move_cost)
                return max_moves
            
            else:  # Bob's turn, minimize the moves
                min_moves = float('inf')
                for i in range(num_pawns):
                    if mask & (1 << i):  # 如果第i个pawn还在
                        next_mask = mask ^ (1 << i)
                        move_cost = knight_to_pawn_dist[i] if current_pawn == -1 else pawn_to_pawn_dist[current_pawn][i]
                        min_moves = min(min_moves, dfs(next_mask, 1, i) + move_cost)
                return min_moves
        
        # 初始状态：Alice 先手，当前骑士还没移动，所有pawns都还在
        return dfs((1 << num_pawns) - 1, 1, -1)

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