<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 {
private:
    vector<vector<int>> DIRs = {
        {1,0},{0,1},{-1,0},{0,-1}
    };

    int dist(int x1, int y1, int x2, int y2) {
        return abs(x1 - x2) + abs(y1 - y2);
    }

    struct comp{
        bool operator()(const vector<int>&  v1, const vector<int>& v2) {
            return v1[2] < v2[2];
        }
    };

public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<vector<int>> q;

        vector<vector<int>> dists(n, vector<int>(n, -1));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    dists[i][j] = 0;
                    q.push({i, j});
                }
            }
        }

        while(!q.empty()) {
            int lx = q.front()[0];
            int ly = q.front()[1];
            q.pop();

            for (int k = 0; k < 4; k++) {
                int nx = lx + DIRs[k][0];
                int ny = ly + DIRs[k][1];

                if (nx >= 0 && nx < n&& ny >= 0 && ny < n && dists[nx][ny] == -1) {
                    dists[nx][ny] = dists[lx][ly] + 1;
                    q.push({nx, ny});
                }
            }
        }

        vector<vector<int>> sfs(n, vector<int>(n, -1));
        // priority_queue<vector<int>, vector<vector<int>>, comp> pq;   // v[0]:x, v[1]:y; v[2]:sf
        priority_queue<vector<int>> pq;   // v[0]:sf, v[1]:x; v[2]:y
        pq.push({dists[0][0], 0, 0});

        while(!pq.empty()) {
            int lsf = pq.top()[0];
            int lx = pq.top()[1];
            int ly = pq.top()[2];
            pq.pop();

            if (sfs[lx][ly] != -1) continue;
            
            sfs[lx][ly] = lsf;
            if (lx == n-1 && ly == n-1) {
                return lsf;
            }

            for (int k = 0; k < 4; k++) {
                int nx = lx + DIRs[k][0];
                int ny = ly + DIRs[k][1];

                if (nx >= 0 && nx < n && ny >= 0 && ny < n && sfs[nx][ny] == -1) {
                    pq.push({min(lsf, dists[nx][ny]), nx, ny});
                }
            }
        }

        return sfs[n-1][n-1];
    }

    // int maximumSafenessFactor(vector<vector<int>>& grid) {
    //     int m = grid.size();
    //     int n = grid[0].size();

    //     vector<vector<int>> thieves;
    //     for (int i = 0; i < m; i++) {
    //         for (int j = 0; j < n; j++) {
    //             if (grid[i][j] == 1) thieves.push_back({i, j});
    //         }
    //     }

    //     vector<vector<int>> dists(m, vector<int>(n, m+n));
    //     int tl = thieves.size();
    //     for (int k = 0; k < tl; k++) {
    //         int tx = thieves[k][0];
    //         int ty = thieves[k][1];

    //         for (int i = 0; i < m; i++) {
    //             for (int j = 0; j < n; j++) {
    //                 if (grid[i][j] == 0) dists[i][j] = min(dists[i][j], dist(tx, ty, i, j));
    //                 else dists[i][j] = 0;
    //             }
    //         }
    //     }

    //     vector<vector<int>> dp(m, vector<int>(n, -1));
    //     queue<vector<int>> q;
    //     q.push({0,0});
    //     dp[0][0] = dists[0][0];

    //     while(!q.empty()) {
    //         int lx = q.front()[0];
    //         int ly = q.front()[1];
    //         q.pop();

    //         for (int k = 0; k < 4; k++) {
    //             int nx = lx + DIRs[k][0];
    //             int ny = ly + DIRs[k][1];

    //             if (nx >= 0 && nx < m && ny >= 0 && ny < n) {
    //                 // if (grid[nx][ny] == 1 && dp[nx][ny] == -1) dp[nx][ny] = 0;
    //                 // else 
    //                 if (dp[nx][ny] == -1 || dp[nx][ny] < min(dists[nx][ny], dp[lx][ly])) {
    //                     dp[nx][ny] = min(dists[nx][ny], dp[lx][ly]);
    //                     q.push({nx, ny});
    //                 }
    //             }
    //         }
    //     }

    //     return dp[m-1][n-1];
    // }
};
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