<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:
        DIR = [(-2,-1),(-2,1),(-1,-2),(-1,2),(1,-2),(1,2),(2,-1),(2,1)]
        all_dists = defaultdict(dict)
        inf = float('inf')
        def get_dists(sx, sy):
            nonlocal all_dists
            dists = {(sx, sy): 0}
        
            q = deque([(sx, sy)])
            while q:
                x,y = q.popleft()
                for dx, dy in DIR:
                    nx, ny = x + dx, y + dy
                    if nx >= 0 and ny >= 0 and nx < 50 and ny < 50 and (nx, ny) not in dists:
                        dists[nx, ny] = 1 + dists[x, y]
                        q.append((nx, ny))

            all_dists[sx, sy] = dists

        # O(len(positions)) * O(V + E)
        # O(15+1) * O((8+1) * 50 * 50)
        # ~O(360,000) to generate all dists
        for sx,sy in positions + [(kx, ky)]:
            get_dists(sx, sy)

        # O(2^n) <= O(32768)
        n = len(positions)
        DONE = (1 << n) - 1

        # best = {}
        @cache
        def dp_best(x, y, state):
            # if (x, y, state) in best and score <= best[x, y, state]:
            #     return 0
            # best[x,y,state] = score

            if state == DONE:
                return 0
            
            res = 0
            for i in range(n):
                if state & (1<<i) == 0:
                    px, py = positions[i]
                    steps = all_dists[x,y][px, py]

                    # We want to maximize the score
                    res = max(res, steps + dp_worst(px, py, state | (1 << i)))
                    # print(px, py, nextscore)
            
            # best[x, y, state] = res
            return res

        # worst = {}
        @cache
        def dp_worst(x, y, state):
            # if (x, y, state) in worst and score >= worst[x, y, state]:
            #     return 0
            # worst[x,y,state] = score
            if state == DONE:
                return 0

            res = inf
            for i in range(n):
                if state & (1<<i) == 0:
                    px, py = positions[i]
                    steps = all_dists[x,y][px, py]

                    # We want to minimize the score
                    res = min(res, steps + dp_best(px, py, state | (1 << i)))
            
            # worst_cache[x, y, state] = res
            return res

        
        # print(positions)
        # print(best)
        # print(worst)
            
        return dp_best(kx, ky, 0)
            
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