<p>You are given a <strong>0-indexed</strong> 2D matrix <code>grid</code> of size <code>n x n</code>, where <code>(r, c)</code> represents:</p>

<ul>
	<li>A cell containing a thief if <code>grid[r][c] = 1</code></li>
	<li>An empty cell if <code>grid[r][c] = 0</code></li>
</ul>

<p>You are initially positioned at cell <code>(0, 0)</code>. In one move, you can move to any adjacent cell in the grid, including cells containing thieves.</p>

<p>The <strong>safeness factor</strong> of a path on the grid is defined as the <strong>minimum</strong> manhattan distance from any cell in the path to any thief in the grid.</p>

<p>Return <em>the <strong>maximum safeness factor</strong> of all paths leading to cell </em><code>(n - 1, n - 1)</code><em>.</em></p>

<p>An <strong>adjacent</strong> cell of cell <code>(r, c)</code>, is one of the cells <code>(r, c + 1)</code>, <code>(r, c - 1)</code>, <code>(r + 1, c)</code> and <code>(r - 1, c)</code> if it exists.</p>

<p>The <strong>Manhattan distance</strong> between two cells <code>(a, b)</code> and <code>(x, y)</code> is equal to <code>|a - x| + |b - y|</code>, where <code>|val|</code> denotes the absolute value of val.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example1.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All paths from (0, 0) to (n - 1, n - 1) go through the thieves in cells (0, 0) and (n - 1, n - 1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example2.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1],[0,0,0],[0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 2) is cell (0, 0). The distance between them is | 0 - 0 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example3.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 3) is cell (1, 2). The distance between them is | 0 - 1 | + | 3 - 2 | = 2.
- The closest cell of the path to the thief at cell (3, 0) is cell (3, 2). The distance between them is | 3 - 3 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

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

<ul>
	<li><code>1 &lt;= grid.length == n &lt;= 400</code></li>
	<li><code>grid[i].length == n</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one thief in the <code>grid</code>.</li>
</ul>

from collections import deque
from heapq import heappush, heappop

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        """
        Idea:
        First, run BFS to mark cells shortest distance to the nearest 1.
        Second, run Dijkstra's algorithm to find the max safeness factor.
        """
        
        L = len(grid)
        
        dist = [[-1] * L for _ in range(L)]
        queue = deque()    # (r, c)
        for r in range(L):
            for c in range(L):
                if grid[r][c] == 1:
                    queue.append((r, c))
                    dist[r][c] = 0
        
        steps = 0
        while queue:
            for _ in range(len(queue)):
                r, c = queue.popleft()
                # dist[r][c] = steps
                
                for dx, dy in [[-1, 0], [1, 0], [0, -1], [0, 1]]:
                    new_r, new_c = r + dx, c + dy
                    if 0 <= new_r < L and 0 <= new_c < L and dist[new_r][new_c] == -1:
                        dist[new_r][new_c] = steps + 1
                        queue.append((new_r, new_c))
            steps += 1
        
        max_heap = [(-dist[0][0], (0, 0))]    # (shortest dist to nearest 1, (r, c))
        visited = [[False] * L for _ in range(L)]
        visited[0][0] = True
        while max_heap:
            d, (r, c) = heappop(max_heap)
            d = -d
            
            if (r, c) == (L - 1, L - 1):
                return d
            
            # if visited[r][c]:
            #     continue
            # visited[r][c] = True
            
            for dx, dy in [[-1, 0], [1, 0], [0, -1], [0, 1]]:
                new_r, new_c = r + dx, c + dy
                if 0 <= new_r < L and 0 <= new_c < L and not visited[new_r][new_c]:
                    visited[new_r][new_c] = True
                    new_d = min(dist[new_r][new_c], d)
                    heappush(max_heap, (-new_d, (new_r, new_c)))
        
        # return -1
3327