<p>You are given a <strong>0-indexed</strong> <code>m x n</code> integer matrix <code>grid</code>. Your initial position is at the <strong>top-left</strong> cell <code>(0, 0)</code>.</p>

<p>Starting from the cell <code>(i, j)</code>, you can move to one of the following cells:</p>

<ul>
	<li>Cells <code>(i, k)</code> with <code>j &lt; k &lt;= grid[i][j] + j</code> (rightward movement), or</li>
	<li>Cells <code>(k, j)</code> with <code>i &lt; k &lt;= grid[i][j] + i</code> (downward movement).</li>
</ul>

<p>Return <em>the minimum number of cells you need to visit to reach the <strong>bottom-right</strong> cell</em> <code>(m - 1, n - 1)</code>. If there is no valid path, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/25/ex1.png" style="width: 271px; height: 171px;" />
<pre>
<strong>Input:</strong> grid = [[3,4,2,1],[4,2,3,1],[2,1,0,0],[2,4,0,0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The image above shows one of the paths that visits exactly 4 cells.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/25/ex2.png" style="width: 271px; height: 171px;" />
<pre>
<strong>Input:</strong> grid = [[3,4,2,1],[4,2,1,1],[2,1,1,0],[3,4,1,0]]
<strong>Output:</strong> 3
<strong>Explanation: </strong>The image above shows one of the paths that visits exactly 3 cells.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/26/ex3.png" style="width: 181px; height: 81px;" />
<pre>
<strong>Input:</strong> grid = [[2,1,0],[1,0,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It can be proven that no path exists.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= grid[i][j] &lt; m * n</code></li>
	<li><code>grid[m - 1][n - 1] == 0</code></li>
</ul>

class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        col_heaps = [[] for _ in grid[0]]  # 每一列的最小堆
        for i, row in enumerate(grid):
            row_h = []  # 第 i 行的最小堆
            for j, (g, col_h) in enumerate(zip(row, col_heaps)):
                while row_h and row_h[0][1] < j:  # 无法到达第 j 列
                    heappop(row_h)  # 弹出无用数据
                while col_h and col_h[0][1] < i:  # 无法到达第 i 行
                    heappop(col_h)  # 弹出无用数据
                f = inf if i or j else 1  # 起点算 1 个格子
                if row_h: f = row_h[0][0] + 1  # 从左边跳过来
                if col_h: f = min(f, col_h[0][0] + 1)  # 从上边跳过来
                if g and f < inf:
                    heappush(row_h, (f, g + j))  # 经过的格子数，向右最远能到达的列号
                    heappush(col_h, (f, g + i))  # 经过的格子数，向下最远能到达的行号
        return f if f < inf else -1  # 此时的 f 是在 (m-1, n-1) 处算出来的

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