<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:
    bool check(int x, int y, int n, int m) {
        return 0 <= x && x < n && 0 <= y && y < m;
    }

    int shortest_path(const vector<vector<int>>& forest, int sx, int sy, int tx, int ty) {

        for(auto& row: dist) {
            for(int i = 0; i < row.size(); ++i) {
                row[i] = -1;
            }
        }
        dist[sx][sy] = 0;

        queue<pair<int, int>> q;
        q.push({sx, sy});

        while (!q.empty()) {
            auto pr = q.front();
            q.pop();
            // cout << pr.first << " " << pr.second << " " << dist[pr.first][pr.second] << endl;
            if (pr.first == tx && pr.second == ty) {
                return dist[tx][ty];
            }

            for(int i = 0; i < 4; ++i) {
                if (check(pr.first + dx[i], pr.second + dy[i], forest.size(), forest[0].size())) {
                    if (dist[pr.first + dx[i]][pr.second + dy[i]] == -1 && forest[pr.first + dx[i]][pr.second + dy[i]] != 0) {
                        q.push({pr.first + dx[i], pr.second + dy[i]});
                        dist[pr.first + dx[i]][pr.second + dy[i]] = dist[pr.first][pr.second] + 1;
                    }
                }
            }
        }
        return dist[tx][ty];
    }

    vector<pair<int, pair<int, int>>> trees;

    int cutOffTree(vector<vector<int>>& forest) {
        dist = vector<vector<int>>(forest.size(), vector<int>(forest[0].size(), 0));

        for(int i = 0; i < forest.size(); ++i) {
            for(int j = 0; j < forest[0].size(); ++j) {
                if (forest[i][j] > 1) {
                    trees.emplace_back(forest[i][j], pair(i, j));
                }
            }
        }
        std::sort(trees.begin(), trees.end());
        int prevx = 0, prevy = 0;
        int result = 0;
        for(int i = 0; i < trees.size(); ++i) {
            int path_length = shortest_path(forest, prevx, prevy, trees[i].second.first, trees[i].second.second);
            // cout << prevx << " " << prevy << " " << trees[i].second.first << " " <<  trees[i].second.second << " " << path_length << endl;
            if (path_length == -1) {
                return -1;
            }
            prevx = trees[i].second.first;
            prevy = trees[i].second.second;
            result += path_length;
        }

        return result;
    }

private:
    vector<vector<int>> dist;
    vector<int> dx = {0, 1, 0, -1};
    vector<int> dy = {1, 0, -1, 0};
};
119045