<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 {
    int rows, cols;
    vector<vector<bool>> not_visited;

  public:
    int cutOffTree(const vector<vector<int>> &forest) {
        rows = forest.size(), cols = forest.front().size();

        vector<pair<int, pair<int, int>>> trees;
        for (int i = 0; i < rows; i++)
            for (int j = 0; j < cols; j++)
                if (forest[i][j] > 1)
                    trees.push_back({forest[i][j], {i, j}});

        sort(trees.begin(), trees.end());

        int r = 0, c = 0;

        int steps = 0;
        for (const pair<int, pair<int, int>> &p : trees) {
            priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<>>
                pq;

            not_visited = vector<vector<bool>>(rows, vector<bool>(cols, true));

            pq.push({0, {r, c}});
            not_visited[r][c] = false;
            bool flag = false;

            while (!pq.empty()) {
                const pair<int, pair<int, int>> pp = pq.top();
                pq.pop();

                if (pp.second == p.second) {
                    flag = true;
                    steps += pp.first;
                    r = p.second.first, c = p.second.second;

                    break;
                }

                for (int i = -1; i <= 1; i += 2) {
                    const int rr = pp.second.first + i, cc = pp.second.second;

                    if (rr >= 0 && rr < rows && not_visited[rr][cc] && forest[rr][cc] >= 1)
                        not_visited[rr][cc] = false, pq.push({pp.first + 1, {rr, cc}});
                }

                for (int i = -1; i <= 1; i += 2) {
                    const int rr = pp.second.first, cc = pp.second.second + i;

                    if (cc >= 0 && cc < cols && not_visited[rr][cc] && forest[rr][cc] >= 1)
                        not_visited[rr][cc] = false, pq.push({pp.first + 1, {rr, cc}});
                }
            }

            if (!flag)
                return -1;
        }

        return steps;
    }
};

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