<p>You are asked to cut off all the trees in a forest for a golf event. The forest is represented as an <code>m x n</code> matrix. In this matrix:</p>

<ul>
	<li><code>0</code> means the cell cannot be walked through.</li>
	<li><code>1</code> represents an empty cell that can be walked through.</li>
	<li>A number greater than <code>1</code> represents a tree in a cell that can be walked through, and this number is the tree&#39;s height.</li>
</ul>

<p>In one step, you can walk in any of the four directions: north, east, south, and west. If you are standing in a cell with a tree, you can choose whether to cut it off.</p>

<p>You must cut off the trees in order from shortest to tallest. When you cut off a tree, the value at its cell becomes <code>1</code> (an empty cell).</p>

<p>Starting from the point <code>(0, 0)</code>, return <em>the minimum steps you need to walk to cut off all the trees</em>. If you cannot cut off all the trees, return <code>-1</code>.</p>

<p><strong>Note:</strong> The input is generated such that no two trees have the same height, and there is at least one tree needs to be cut off.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/trees1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> forest = [[1,2,3],[0,0,4],[7,6,5]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Following the path above allows you to cut off the trees from shortest to tallest in 6 steps.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/trees2.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> forest = [[1,2,3],[0,0,0],[7,6,5]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> The trees in the bottom row cannot be accessed as the middle row is blocked.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> forest = [[2,3,4],[0,0,5],[8,7,6]]
<strong>Output:</strong> 6
<b>Explanation:</b> You can follow the same path as Example 1 to cut off all the trees.
Note that you can cut off the first tree at (0, 0) before making any steps.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == forest.length</code></li>
	<li><code>n == forest[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>0 &lt;= forest[i][j] &lt;= 10<sup>9</sup></code></li>
	<li>Heights of all trees are <strong>distinct</strong>.</li>
</ul>

class Solution {
public:
    int cutOffTree(vector<vector<int>> const& forest) const {
        if (forest[0][0] == 0) return -1;
        int m = forest.size(), n = forest[0].size();
        vector<pair<int, int>> locs;
        for (int i{}; i < m; ++i) {
            for (int j{}; j < n; ++j) {
                if (forest[i][j] > 1) locs.emplace_back(forest[i][j], i*n+j);
            }
        }
        sort(locs.begin(), locs.end(), [](auto const& a, auto const& b){return a.first < b.first;});
        int ans{};
        int cur{0};
        for (auto const& p : locs) {
            int dist{get_dist(forest, cur, p.second)};
            if (dist == INT_MAX) return -1;
            ans += dist;
            cur = p.second;
        }
        return ans;
    }
private:
    static int constexpr ds[][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    int get_dist(vector<vector<int>> const& forest, int s, int d) const {
        if (s == d) return 0;
        queue<int> q;
        q.push(s);
        int ans{1};
        int m = forest.size(), n = forest[0].size();
        vector<bool> visited(m*n);
        visited[s] = true;
        while (not q.empty()) {
            for (int i = q.size(); i > 0; --i) {
                int top{q.front()};
                q.pop();
                int y{top / n}, x{top % n};
                for (auto [dy, dx] : ds) {
                    int ny{y + dy}, nx{x + dx};
                    if (ny < 0 or ny >= m or nx < 0 or nx >= n or forest[ny][nx] == 0 or visited[ny*n+nx]) continue;
                    if (ny*n+nx == d) return ans;
                    q.push(ny*n+nx);
                    visited[ny*n+nx] = true;
                }
            }
            ++ans;
        }
        return INT_MAX;
    }
};
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