<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>


NN = 50
MM = NN - 1
dist = [[[[inf] * NN for i in range(NN)] for j in range(NN)] for k in range(NN)]

for x1 in range(NN // 2):
    for y1 in range(x1 + 1):
        queue = [[x1, y1, 0]]
        dist[x1][y1][x1][y1] = 0
        for x, y, d in queue:
            for nx, ny in [
                [x - 2, y + 1],
                [x - 2, y - 1],
                [x + 2, y + 1],
                [x + 2, y - 1],
                [x - 1, y + 2],
                [x - 1, y - 2],
                [x + 1, y + 2],
                [x + 1, y - 2],
            ]:
                if 0 <= nx < NN > ny >= 0 and dist[x1][y1][nx][ny] == inf:
                    dist[x1][y1][nx][ny] = d + 1
                    queue.append([nx, ny, d + 1])

        x1m = MM - x1
        y1m = MM - y1

        for x2, y2 in product(range(NN), repeat=2):
            d = dist[x1][y1][x2][y2]
            x2m = MM - x2
            y2m = MM - y2

            dist[y1][x1][y2][x2] = d

            dist[x1m][y1][x2m][y2] = d
            dist[x1][y1m][x2][y2m] = d
            dist[x1m][y1m][x2m][y2m] = d

            dist[y1m][x1][y2m][x2] = d
            dist[y1][x1m][y2][x2m] = d
            dist[y1m][x1m][y2m][x2m] = d



fmin = lambda x, y: x if x < y else y
fmax = lambda x, y: x if x > y else y
class Solution:
    def maxMoves(self, kx: int, ky: int, A: List[List[int]]) -> int:
        R = C = 50
        N = len(A)

        dp = [[0] * N for _ in range(1 << N)]
        for mask in range(1 << N):
            count = bin(mask).count("1")
            for i in range(N):
                dp[mask][i] = inf if count % 2 == 1 else -inf
        dp[(1 << N) - 1] = [0] * N

        for nmask in range((1 << N) - 1, 0, -1):
            count = bin(nmask).count("1")
            fn = fmin if count % 2 == 0 else fmax

            for v, (xnode, ynode) in enumerate(A):
                if nmask >> v & 1:
                    pmask = nmask ^ (1 << v)
                    for u, (xnei, ynei) in enumerate(A):
                        if pmask >> u & 1:
                            dp[pmask][u] = fn(
                                dp[pmask][u],
                                dp[nmask][v] + dist[xnode][ynode][xnei][ynei],
                            )

        # fn = min if N % 2 else max
        ans = 0
        for first, (x, y) in enumerate(A):
            d = dp[1 << first][first]
            d += dist[x][y][kx][ky]
            ans = fmax(ans, d)
        return ans

90471