<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])
        pre = []
        for i in range(m):
            for j in range(n):
                pre.append([board[i][j], i, j])
        pre.sort(reverse=True)
        t = m * n

        def cal2(x1, y1, x2, y2):
            for i in range(t):
                if pre[i][1] != x1 and pre[i][1] != x2 and pre[i][2] != y1 and pre[i][2] != y2:
                    return i

        def cal1(x1, y1):
            for i in range(t):
                if pre[i][1] != x1 and pre[i][2] != y1:
                    return i

        res = -inf
        x1 = pre[0][1]
        for y1 in range(n):
            p1=board[x1][y1]
            k = cal1(x1, y1)
            y2 = pre[k][2]
            for x2 in range(m):
                if x2==x1:
                    continue
                p2= p1+board[x2][y2]
                r = pre[cal2(x1, y1, x2, y2)][0]
                p3=p2+r
                if p3>res:
                    res=p3
            x2 = pre[k][1]
            for y2 in range(n):
                if y2==y1:
                    continue
                p2= p1+board[x2][y2]
                r = pre[cal2(x1, y1, x2, y2)][0]
                p3=p2+r
                if p3>res:
                    res=p3

        y1 = pre[0][2]
        for x1 in range(m):
            p1=board[x1][y1]
            k = cal1(x1, y1)
            y2 = pre[k][2]
            for x2 in range(m):
                if x2==x1:
                    continue
                p2= p1+board[x2][y2]
                r = pre[cal2(x1, y1, x2, y2)][0]
                p3=p2+r
                if p3>res:
                    res=p3
            x2 = pre[k][1]
            for y2 in range(n):
                if y2==y1:
                    continue
                p2= p1+board[x2][y2]
                r = pre[cal2(x1, y1, x2, y2)][0]
                p3=p2+r
                if p3>res:
                    res=p3
        return res



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