Maximum Number of Moves to Kill All Pawns
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3283.Maximum%20Number%20of%20Moves%20to%20Kill%20All%20Pawns/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3283.Maximum%20Number%20of%20Moves%20to%20Kill%20All%20Pawns/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3283.Maximum%20Number%20of%20Moves%20to%20Kill%20All%20Pawns/images/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>
Bit Manipulation; Breadth-First Search; Array; Math; Bitmask; Game Theory
class Solution:
    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        @cache
        def dfs(last: int, state: int, k: int) -> int:
            if state == 0:
                return 0
            if k:
                res = 0
                for i, (x, y) in enumerate(positions):
                    if state >> i & 1:
                        t = dfs(i, state ^ (1 << i), k ^ 1) + dist[last][x][y]
                        if res < t:
                            res = t
                return res
            else:
                res = inf
                for i, (x, y) in enumerate(positions):
                    if state >> i & 1:
                        t = dfs(i, state ^ (1 << i), k ^ 1) + dist[last][x][y]
                        if res > t:
                            res = t
                return res

        n = len(positions)
        m = 50
        dist = [[[-1] * m for _ in range(m)] for _ in range(n + 1)]
        dx = [1, 1, 2, 2, -1, -1, -2, -2]
        dy = [2, -2, 1, -1, 2, -2, 1, -1]
        positions.append([kx, ky])
        for i, (x, y) in enumerate(positions):
            dist[i][x][y] = 0
            q = deque([(x, y)])
            step = 0
            while q:
                step += 1
                for _ in range(len(q)):
                    x1, y1 = q.popleft()
                    for j in range(8):
                        x2, y2 = x1 + dx[j], y1 + dy[j]
                        if 0 <= x2 < m and 0 <= y2 < m and dist[i][x2][y2] == -1:
                            dist[i][x2][y2] = step
                            q.append((x2, y2))

        ans = dfs(n, (1 << n) - 1, 1)
        dfs.cache_clear()
        return ans
