<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 DSU {
public:
    DSU(int n) {
        prnt = vector<int>(n + 1);
        val = vector<int>(n + 1);
        for (int i = 1; i <= n; i++) {
            prnt[i] = i;
            val[i] = 1;
        }
    }
    int root(int x) {
        if (x != prnt[x]) {
            return prnt[x] = root(prnt[x]);
        }
        return x;
    }

    bool merge(int x, int y) {
        int px = root(x), py = root(y);
        if (px == py) {
            return false;
        }
        if (val[px] > val[py]) {
            swap(py, px);
        }
        prnt[px] = py;
        val[py] += val[px];
        return true;
    }
    vector<int> prnt, val;
};

class Solution {
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> dis(n, vector<int>(n, 1e9));
        queue<pair<int, int>> q;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j]) {
                    dis[i][j] = 0;
                    q.push({i, j});
                }
            }
        }
        auto valid = [&](int x, int y) -> bool {
            if (x >= 0 && y >= 0 && x < n && y < n && dis[x][y] == 1e9) {
                return true;
            }
            return false;
        };

        while (!q.empty()) {
            int x = q.front().first, y = q.front().second;
            q.pop();
            if (valid(x, y + 1)) {
                q.push({x, y + 1});
                dis[x][y + 1] = dis[x][y] + 1;
            }
            if (valid(x, y - 1)) {
                q.push({x, y - 1});
                dis[x][y - 1] = dis[x][y] + 1;
            }
            if (valid(x + 1, y)) {
                q.push({x + 1, y});
                dis[x + 1][y] = dis[x][y] + 1;
            }
            if (valid(x - 1, y)) {
                q.push({x - 1, y});
                dis[x - 1][y] = dis[x][y] + 1;
            }
        }

        vector<vector<int>> all;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                all.push_back({dis[i][j], i, j});
            }
        }
        sort(all.begin(), all.end());
        reverse(all.begin(), all.end());
        DSU dsu(n * n);
        map<pair<int, int>, bool> appeared;
        const int last = n * n - 1;
        for (auto v : all) {
            int d = v[0], x = v[1], y = v[2];
            appeared[{x, y}] = true;
            int idx = x * n + y;
            if (appeared[{x, y + 1}]) {
                dsu.merge(idx, idx + 1);
            }
            if (appeared[{x, y - 1}]) {
                dsu.merge(idx, idx - 1);
            }
            if (appeared[{x - 1, y}]) {
                dsu.merge(idx, idx - n);
            }
            if (appeared[{x + 1, y}]) {
                dsu.merge(idx, idx + n);
            }
            if (dsu.root(0) == dsu.root(last)) {
                return d;
            }
        }
        return -1;
    }
};
405121