<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 dx[4] = {0,0,-1,1};
    int dy[4] = {-1,1,0,0};
    int cutOffTree(vector<vector<int>>& forest) {
        int n = (int)forest.size();
        int m = (int)forest[0].size();
        vector<array<int, 3>> sorted;
        int ans = 0;
        for (int i = 0; i<n; ++i)
            for (int j = 0; j<m; ++j)
                if (forest[i][j] > 1)
                    sorted.push_back({forest[i][j], i, j});
        sort(sorted.begin(), sorted.end());

        int x = 0;
        int y = 0;
        vector<vector<int>> seen(n, vector<int>(m, -1));
        for (int i = 0; i<(int)sorted.size(); ++i) {
            int lookx = sorted[i][1];
            int looky = sorted[i][2];
            vector<pair<int, int>> stack;
            stack.push_back({x,y});
            seen[x][y] = i;
            bool found = false;
            int c = 0;
            while (!stack.empty() && !found) {
                vector<pair<int, int>> nextstack;
                int sz = (int)stack.size();
                for (int j = 0; j<sz; ++j) {
                    int currentx = stack[j].first;
                    int currenty = stack[j].second;
                    if (currentx == lookx && currenty == looky) {
                        found = true;
                        break;
                    }
                    for (int p = 0; p<4; ++p) {
                        int _x = dx[p] + currentx;
                        int _y = dy[p] + currenty;
                        if (_x >= 0 && _x < n && _y >=0 && _y<m && seen[_x][_y] != i && forest[_x][_y] > 0) {
                            seen[_x][_y] = i;
                            nextstack.push_back({_x, _y});
                        }
                    }
                }
                if (found) break;
                swap(stack, nextstack);
                c++;
            }

            if (!found)
                return -1;
            else {
                ans += c;
                x = lookx;
                y = looky;
            }
        }

        return ans;



        return ans;





    }
};
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