<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:
        
        def encode(r, c):
            return r * 50 + c
        
        def decode(x):
            return x // 50, x % 50
        
        positions.append([kx, ky])
                
        dr = [-2, -1, 1, 2, 2, 1, -1, -2]
        dc = [1, 2, 2, 1, -1, -2, -2, -1]
        
        def bfs(x, y):
            maxD = int(1e5)        
            dd = [maxD for _ in range(2505)]
            dd[x] = 0
            q = deque()
            q.append(x)
            while q:
                focus = q.popleft()
                if focus == y:
                    return dd[focus]
                r, c = decode(focus)
                for j in range(len(dr)):
                    nr = r + dr[j]
                    nc = c + dc[j]
                    if nr < 0 or nr >= 50 or nc < 0 or nc >= 50:
                        continue
                    nxt = encode(nr, nc)
                    if dd[focus] + 1 < dd[nxt]:
                        q.append(nxt)
                        dd[nxt] = dd[focus] + 1
            return dd[y]
        
        n = len(positions)
        dist = [[0 for _ in range(n)] for _ in range(n)]
        
        for i in range(n):
            [a, b] = positions[i]
            x = encode(a, b)
            for j in range(n):
                [c, d] = positions[j]
                y = encode(c, d)
                dist[i][j] = bfs(x, y)
        
        dpmax = [[-1 for _ in range(1 << n)] for _ in range(n)]
        dpmin = [[float('inf') for _ in range(1 << n)] for _ in range(n)]
        
        for i in range(n):
            dpmin[i][0] = 0
            dpmax[i][0] = 0
            
        b = {}
        for bitmask in range(1 << (n)):
            r = []
            for i in range(n):
                if ((bitmask >> i) & 1) == 1:
                    r.append(i)
            b[bitmask] = r
        
        
        def dfs(bitmask, location, alice):
            nonlocal n
            if alice:
                if dpmax[location][bitmask] != -1:
                    return dpmax[location][bitmask]
                for i in b[bitmask]:
                    # print(bitmask, i, (bitmask >> i))
                    dpmax[location][bitmask] = max(dpmax[location][bitmask], 
                                                       dfs(bitmask ^ (1 << i), i, False) + dist[i][location])
                return dpmax[location][bitmask]
            else:
                if dpmin[location][bitmask] < float('inf') / 2:
                    return dpmin[location][bitmask]
                for i in b[bitmask]:
                    dpmin[location][bitmask] = min(dpmin[location][bitmask], 
                                                       dfs(bitmask ^ (1 << i), i, True) + dist[i][location])
            
            
                return dpmin[location][bitmask]
        
        dfs((1 << (n - 1)) - 1, n - 1, True)
        
        
        return dpmax[n - 1][(1 << (n - 1)) - 1]


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