<p>You are given a <code>m x n</code> 2D array <code>board</code> representing a chessboard, where <code>board[i][j]</code> represents the <strong>value</strong> of the cell <code>(i, j)</code>.</p>

<p>Rooks in the <strong>same</strong> row or column <strong>attack</strong> each other. You need to place <em>three</em> rooks on the chessboard such that the rooks <strong>do not</strong> <strong>attack</strong> each other.</p>

<p>Return the <strong>maximum</strong> sum of the cell <strong>values</strong> on which the rooks are placed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = </span>[[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]</p>

<p><strong>Output:</strong> 4</p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/08/rooks2.png" style="width: 294px; height: 450px;" /></p>

<p>We can place the rooks in the cells <code>(0, 2)</code>, <code>(1, 3)</code>, and <code>(2, 1)</code> for a sum of <code>1 + 1 + 2 = 4</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[1,2,3],[4,5,6],[7,8,9]]</span></p>

<p><strong>Output:</strong> <span class="example-io">15</span></p>

<p><strong>Explanation:</strong></p>

<p>We can place the rooks in the cells <code>(0, 0)</code>, <code>(1, 1)</code>, and <code>(2, 2)</code> for a sum of <code>1 + 5 + 9 = 15</code>.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[1,1,1],[1,1,1],[1,1,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>We can place the rooks in the cells <code>(0, 2)</code>, <code>(1, 1)</code>, and <code>(2, 0)</code> for a sum of <code>1 + 1 + 1 = 3</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= m == board.length &lt;= 500</code></li>
	<li><code>3 &lt;= n == board[i].length &lt;= 500</code></li>
	<li><code>-10<sup>9</sup> &lt;= board[i][j] &lt;= 10<sup>9</sup></code></li>
</ul>


class Solution:
    def maximumValueSum(self, board: List[List[int]]) -> int:
        m, n = len(board), len(board[0])
        loc = sorted([(board[i][j], i, j) for i in range(m) for j in range(n)], reverse=True)  # 每个价值及其出现的位置，按价值降序排列
        row = set()  # 已选中的行
        col = set()  # 已选中的列
        total = 0  # 已选中的价值和
        ans = - maxsize

        # 从loc[x:]中选一个位置
        def dfs(x):
            nonlocal total, ans
            for y, (val, i, j) in enumerate(loc[x:], start=x):
                if i not in row and j not in col:
                    # 之后的价值和更小，剪枝
                    if len(row) == 2:
                        if total + val > ans:
                            ans = total + val
                        return
                    # 价值和不可能超过ans，剪枝
                    if (len(row) == 1 and total + val * 2 <= ans) or (len(row) == 0 and val * 3 <= ans):
                        return
                    total += val
                    row.add(i)
                    col.add(j)
                    dfs(y + 1)
                    total -= val
                    row.remove(i)
                    col.remove(j)
        
        dfs(0)
        return ans



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