Find the Safest Path in a Grid
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2812.Find%20the%20Safest%20Path%20in%20a%20Grid/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2812.Find%20the%20Safest%20Path%20in%20a%20Grid/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2812.Find%20the%20Safest%20Path%20in%20a%20Grid/images/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>
Breadth-First Search; Union Find; Array; Binary Search; Matrix; Heap (Priority Queue)
class UnionFind {
public:
    vector<int> p;
    int n;

    UnionFind(int _n)
        : n(_n)
        , p(_n) {
        iota(p.begin(), p.end(), 0);
    }

    bool unite(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) return false;
        p[pa] = pb;
        --n;
        return true;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

class Solution {
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        if (grid[0][0] || grid[n - 1][n - 1]) {
            return 0;
        }
        queue<pair<int, int>> q;
        int dist[n][n];
        memset(dist, 0x3f, sizeof(dist));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j]) {
                    dist[i][j] = 0;
                    q.emplace(i, j);
                }
            }
        }
        int dirs[5] = {-1, 0, 1, 0, -1};
        while (!q.empty()) {
            auto [i, j] = q.front();
            q.pop();
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && dist[x][y] == 0x3f3f3f3f) {
                    dist[x][y] = dist[i][j] + 1;
                    q.emplace(x, y);
                }
            }
        }
        vector<tuple<int, int, int>> t;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                t.emplace_back(dist[i][j], i, j);
            }
        }
        sort(t.begin(), t.end());
        reverse(t.begin(), t.end());
        UnionFind uf(n * n);
        for (auto [d, i, j] : t) {
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && dist[x][y] >= d) {
                    uf.unite(i * n + j, x * n + y);
                }
            }
            if (uf.find(0) == uf.find(n * n - 1)) {
                return d;
            }
        }
        return 0;
    }
};