<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>

# distance in general, assuming you can go "out-of-bound"
distance = [[math.inf] * 50 for _ in range(50)]
distance[0][0] = 0
distance[1][2] = 1
distance[2][1] = 1
distance[1][1] = 2
distance[1][3] = 2
distance[3][1] = 2
distance[3][3] = 2
distance[2][0] = 2
distance[4][0] = 2
distance[4][2] = 2
distance[0][2] = 2
distance[0][4] = 2
distance[2][4] = 2

queue = [(1, 1), (1, 3), (3, 1), (3, 3), (2, 0), (4,0), (4, 2), (0, 2), (0, 4), (2, 4)]
steps = 2
while queue:
    next_queue = []
    for x, y in queue:
        for dx, dy in [(1, 2), (1, -2), (-1, 2), (-1, -2), (2, 1), (2, -1), (-2, 1), (-2, -1)]:
            nx, ny = x+dx, y+dy
            if 0<=nx<50 and 0<=ny<50 and distance[nx][ny] > steps+1:
                distance[nx][ny] = steps+1
                next_queue.append((nx, ny))
    queue = next_queue
    steps += 1

class Solution:
    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        n = len(positions)
        END = (2**n)-1
        positions.append((kx, ky)) # register initial position as virtual pawn. only included in distance calculation. not included in usage/mask
        initial_res = [0, math.inf] # Alice, Bob
        op = [max, min] # Alice, Bob
        corner_set = {(0, 0), (0, 49), (49, 9), (49, 49)}
        distance_dp = [[0] * (n+1) for _ in range(n+1)]
        for i in range(n+1):
            x, y = positions[i]
            i_is_corner = (x, y) in corner_set
            for j in range(i+1, n+1):
                x2, y2 = positions[j]
                j_is_corner = (x2, y2) in corner_set
                dx, dy = abs(x-x2), abs(y-y2)
                if dx == 1 and dy == 1 and (i_is_corner or j_is_corner):
                    distance_dp[i][j] = 4
                    distance_dp[j][i] = 4
                else:
                    distance_dp[i][j] = distance[dx][dy]
                    distance_dp[j][i] = distance[dx][dy]

        @cache
        def dp(cur, usage):
            # base case
            if usage == END:
                return 0
            
            ptr = usage.bit_count()%2
            res = initial_res[ptr]
            for i in range(n):
                mask = 1 << i
                if (mask&usage) != 0:
                    # the pawn have been killed
                    continue
                d = distance_dp[cur][i]
                res = op[ptr](res, d + dp(i, usage+mask))
            return res
        
        return dp(n, 0)
        
        
1865