<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 heapq import heappush, heappop
class Solution:
    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        h, w = 50, 50
        def getdist(sx, sy):
            d = [[-1] * w for _ in range(h)]
            d[sy][sx] = 0
            q = deque([(sx, sy)])
            while len(q) > 0:
                x, y = q.popleft()
                for dx, dy in [(2, 1), (1, 2), (-1, 2), (-2, 1), (-2, -1), (-1, -2), (1, -2), (2, -1)]:
                    x1, y1 = x + dx, y + dy
                    if 0 <= x1 < w and 0 <= y1 < h and d[y1][x1] < 0:
                        d[y1][x1] = d[y][x] + 1
                        q.append((x1, y1))
            return d
        positions.append((kx, ky))
        n = len(positions)
        dpos = [[0] * n for _ in range(n)]
        for i in range(n):
            d = getdist(positions[i][0], positions[i][1])
            for j in range(n):
                dpos[i][j] = d[positions[j][1]][positions[j][0]]
        #print(f'dpos={dpos}')                
        c2mask = [[] for _ in range(n)]
        def popcount(m):
            ans = 0
            while m > 0:
                m &= m - 1
                ans += 1
            return ans
        for m in range(1 << (n - 1)):
            c2mask[popcount(m)].append(m)
        dp = [[0] * (n - 1) for _ in range(1 << (n - 1))]
        for cnt in range(1, n - 1):
            for m in c2mask[cnt]:
                for i in range(n - 1):
                    if m & (1 << i) == 0:
                        if (n - cnt) % 2 == 1: # alice
                            x = float('-inf')
                            for j in range(n - 1):
                                if m & (1 << j) != 0:
                                    m1 = m ^ (1 << j)
                                    x = max(x, dp[m1][j] + dpos[i][j])
                            dp[m][i] = x
                        else:
                            x = float('inf')
                            for j in range(n - 1):
                                if m & (1 << j) != 0:
                                    m1 = m ^ (1 << j)
                                    x = min(x, dp[m1][j] + dpos[i][j])
                            dp[m][i] = x
        #print(f'dp={dp}')                            
        ans = 0
        for m in c2mask[n - 2]:
            for i in range(n - 1):
                if m & (1 << i) == 0:
                    x = dp[m][i] + dpos[n - 1][i]
                    ans = max(ans, x)
        return ans                    
        
                    
            
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