<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:
    typedef pair<int,pair<int,int>> pii;
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>>man(n,vector<int>(n,INT_MAX));
        queue<vector<int>>q;
        vector<int>dirx = {1,-1,0,0}, diry = {0,0,1,-1};
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    man[i][j]=0;
                    q.push({i,j});
                }
            }
        }

        while(!q.empty()){
            int sz = q.size();
            while(sz--){
                auto v = q.front();
                q.pop();
                for(int k=0;k<4;k++){
                    int x = v[0]+dirx[k], y = v[1]+diry[k];
                    if(x>=0 && y>=0 && x<n && y<n && man[x][y]>1+man[v[0]][v[1]]){
                        q.push({x,y});
                        man[x][y] = man[v[0]][v[1]]+1;
                    }
                }
            }
        }
    
        vector<vector<int>>vis(n,vector<int>(n,0));
        priority_queue<pii>pq;
        vis[0][0]=1;
        pq.push({man[0][0],{0,0}});
        int ans = INT_MAX;
        while(!pq.empty()){
            auto v = pq.top();
            pq.pop();
            int sf = v.first;
            auto p = v.second;
            ans=min(ans,sf);
            if(p.first==n-1 && p.second==n-1)return ans;
            for(int k=0;k<4;k++){
                int x = p.first+dirx[k], y = p.second+diry[k];
                if(x<0 || x>=n || y<0 || y>=n || vis[x][y])continue;
                pq.push({man[x][y],{x,y}});
                vis[x][y]=1;
            }
        }
        return ans;
        // return -1;
    }
};
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