<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:
    void cal_dist(vector<vector<int>>& grid){
        int n = grid.size();
        int di[] = {0,-1,0,1};
        int dj[] = {1,0,-1,0};
        queue<pair<int,int>> queue;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j)
                if(grid[i][j] == 1) queue.push({i,j});
        }

        while(queue.size() > 0){
            auto [cur_i, cur_j] = queue.front(); queue.pop();
            for(int cnt = 0; cnt < 4; ++cnt){
                int I = cur_i + di[cnt], J = cur_j + dj[cnt];
                if(I < 0 || I >= n || J < 0 || J >= n) continue;
                if(grid[I][J] == 0 || grid[I][J] > grid[cur_i][cur_j] + 1){
                    grid[I][J] = grid[cur_i][cur_j] + 1;
                    queue.push({I,J});
                }
            }
        }
    }
    int find_path(vector<vector<int>>& grid){
        priority_queue<tuple<int,int,int>> queue;
        queue.push({grid[0][0],0,0});
        int di[] = {0,-1,0,1};
        int dj[] = {1,0,-1,0};
        int n = grid.size(), min_dist = n;
        vector<vector<bool>> visit(n, vector<bool>(n,false));
        visit[0][0] = true;
        while(queue.size() > 0){
            auto [dist, cur_i, cur_j] = queue.top(); queue.pop();
             min_dist = min(min_dist,grid[cur_i][cur_j]);
            if(cur_i == n-1 && cur_j == n-1) return min_dist-1;
            for(int cnt = 0; cnt < 4; ++cnt){
                int I = cur_i + di[cnt], J = cur_j + dj[cnt];
                if(I < 0 || I >= n || J < 0 || J >= n) continue;
                if(visit[I][J]) continue;               
                queue.push({grid[I][J],I,J});
                 visit[I][J] = true;
            }
        }

        return min_dist-1;
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        if(grid[0][0] || grid[n - 1][n - 1]) return 0;
        cal_dist(grid);
        
        // for(int i = 0; i < n; ++i){
        //     for(int j = 0; j < n; ++j)
        //         cout << " " << grid[i][j];
        //     cout << endl;
        // }
        return find_path(grid);
    }
};
114806