<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 {
public:
    int n;
    vector<vector<int>> g, td, sf;
    vector<vector<int>> dlst = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        g = grid;
        n = g.size();
        td = vector<vector<int>>(n, vector<int>(n, INT_MAX));   // min dist to thief
        sf = vector<vector<int>>(n, vector<int>(n, 0));         // max safeness factor to here
        bfsDistanceToThief();
        dijkstraPathToGoal();
        return sf[n-1][n-1];
    }
    void bfsDistanceToThief(){
        queue<vector<int>> que; // {(r, c), ...}
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(g[i][j]==1){
                    que.push({i, j});
                    td[i][j]=0;
                }
            }
        }
        int step=1;
        while(!que.empty()){
            int sz=que.size();
            while(sz--){
                int cr = que.front()[0];
                int cc = que.front()[1];
                que.pop();
                for(auto &d : dlst){
                    int nr = cr+d[0];
                    int nc = cc+d[1];
                    if(nr<0 || nr>=n || nc<0 || nc>=n || td[nr][nc]<=step) continue;
                    td[nr][nc]=step;
                    que.push({nr, nc});
                }
            }
            step++;
        }
    }
    void dijkstraPathToGoal(){
        priority_queue<vector<int>> pq; // {(min_fs_till_now, r, c), ...}
        if(g[0][0]==0) pq.push({td[0][0], 0, 0});
        while(!pq.empty()){
            int cf = pq.top()[0];
            int cr = pq.top()[1];
            int cc = pq.top()[2];
            pq.pop();
            for(auto &d : dlst){
                int nr = cr + d[0];
                int nc = cc + d[1];
                if(nr<0 || nr>=n || nc<0 || nc>=n) continue;
                int nf = min(cf, td[nr][nc]);
                if(nf<=sf[nr][nc]) continue;
                sf[nr][nc] = nf;
                if(nr==n-1 && nc==n-1) return;
                pq.push({nf, nr, nc});
            }
        }
    }
};
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