<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>

class Solution:
    """
    n >= 1, at least one theif in the grid

    4-connectivity, undirected graph
    n*n environment matrix: 1 contain thief, 0 empty
    path (0,0)->(n-1,n-1): can include thief cell
    safeness factor of path: min |a-x|+|b-y| for any (x,y) in path and any (a,b) theif
    (if contain theif, then safeness is 0)
    return max safeness factor

    multi-source bfs to compute mahattan distance of (i,j) to nearest theif
        time O(n^2), space O(n^2)
    dijkstra, time O(n^2 logn), space O(n^2)
    total complexity: time O(n^2 logn), space O(n^2)
    """
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        n = len(grid)
        # part 1
        # dist from nearest theif
        distance = [[None] * n for _ in range(n)]
        # bfs, initialize with thief
        queue = deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    queue.append((i, j))
                    distance[i][j] = 0
        # queue = deque([(i, j) for i in range(n) for j in range(n) if grid[i][j] == 1])
        depth = 0
        directions = [(-1, 0), (1, 0), (0, 1), (0, -1)]
        while queue:
            qlen = len(queue)
            for _ in range(qlen):
                i, j = queue.popleft()
                for dx, dy in directions:
                    ni, nj = i+dx, j+dy
                    if 0 <= ni < n and 0 <= nj < n and distance[ni][nj] is None:
                        distance[ni][nj] = depth + 1
                        queue.append((ni, nj))
            depth += 1


        # part 2: dijkstra
        # negative safe
        safe = [[1] * n for _ in range(n)]
        # max heap
        heap = [(-distance[0][0], 0,0)]
        safe[0][0] = -distance[0][0]
        while heap:
            cur_safe, i, j = heappop(heap)
            cur_safe = cur_safe
            if i == n-1 and j == n-1:
                return -cur_safe
            if cur_safe > safe[i][j]: continue
            for dx, dy in directions:
                ni, nj = i + dx, j + dy
                if 0 <= ni < n and 0 <= nj < n:
                    nei_safe = max(cur_safe, -distance[ni][nj])
                    if nei_safe < safe[ni][nj]:
                        safe[ni][nj] = nei_safe
                        heappush(heap, (nei_safe, ni, nj))
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