<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:
        """
        minimax
        50*50 fixed chessboard
        15 pawns only interesting

        all path shortest distance?
        generate all s to all t first
        s:  (kx,ky)&positions   16
        d:  positions           15


        lets assume cant greedily pick next move
        
        O(bfs*N all-pairs st + N*2^N caches* N per run)
                                    7*1e6
        """
        INF=10**20
        N=len(positions)
        ds=[
            (-2,1),(-1,2),(2,1),(1,2),
            (2,-1),(1,-2),(-1,-2),(-2,-1)
        ]
        sToD=[None for _ in range(N+1)] #source: [distance to each dest]
        positions=[(y,x) for x,y in positions]
        sources=list(positions)
        sources.append((ky,kx))
        destinations={v:i for i,v in enumerate(positions)}

        def bfs(j,i):
            q=deque([(0,j,i)])
            oj,oi=j,i
            cnt=0
            visited=set()
            res=[None for _ in range(len(destinations))]
            while q and cnt<len(destinations):
                dis,j,i=q.popleft()
                if (j,i) in visited:
                    continue
                visited.add((j,i))
                # print(f"{dis=} {j=} {i=}")
                if (j,i) in destinations:   #can pass by another destination
                    res[destinations[(j,i)]]=dis
                    cnt+=1
                    # continue
                for d in ds:
                    nj=j+d[0]
                    ni=i+d[1]
                    if 0<=nj<50 and 0<=ni<50:
                        q.append((dis+1,nj,ni))
            #print(f"{(oj,oi)=} {res=}")
            return res

        # print(f"{destinations=}")
        # bfs(1,1)
        #print(positions)
        for index,v in enumerate(sources):
            j,i=v
            sToD[index]=bfs(j,i)
        
        """
        remain: set of remaining pawns
        prev:   prev coordinates' index at sources[N+1]
        """
        @cache
        def dfs(prev,remain):    
            if remain==0:
                return 0
            isBob = (N-remain.bit_count())&1  
            #getMin  
            if isBob:
                res=INF
                for j in range(N):
                    if (1<<j)&remain>0:
                        nMask= remain^(1<<j)    #set jth bit to 0
                        dis= sToD[prev][j]
                        res=min(res,dis+dfs(j,nMask))
                return res
            #getMax
            res=0
            for j in range(N):
                if (1<<j)&remain>0:
                    nMask= remain^(1<<j)    #set jth bit to 0
                    dis= sToD[prev][j]
                    res=max(res,dis+dfs(j,nMask))
            return res


        return dfs(N, (1<<N)-1)  


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