<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>>& forest) {
        int m = forest.size();
        int n = forest[0].size();

        vector<pair<int, pair<int, int>>> trees = {{1, {0, 0}}};
        for(int r=0; r<m; ++r)
            for(int c=0; c<n; ++c)
                if(forest[r][c] > 1)
                    trees.push_back({forest[r][c], {r, c}});

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

        int ans = 0;
        vector<int> idxs = {-1, 0, 1, 0, -1};
        for(int t=0; t+1<trees.size(); ++t){
            vector<vector<int>> dis(m, vector<int>(n, -1));
            int i = trees[t].second.first, j = trees[t].second.second;
            int desI = trees[t+1].second.first, desJ = trees[t+1].second.second;

            dis[i][j] = 0;
            queue<pair<int, int>> q;
            q.push({i, j});

            while(!q.empty()){
                int sz = q.size();
                while(sz--){
                    int cI = q.front().first, cJ = q.front().second;

                    q.pop(); 
                    for(int k=0; k<4; ++k){
                        int nI = cI + idxs[k], nJ = cJ + idxs[k+1];
                        if(nI >= 0 && nI < m && nJ >= 0 && nJ < n && dis[nI][nJ] == -1 && forest[nI][nJ]){
                            dis[nI][nJ] = dis[cI][cJ] + 1;
                            q.push({nI, nJ});
                        }
                    }
                }
            }

            if(dis[desI][desJ] == -1) return -1;
            else ans += dis[desI][desJ];
        }  

        return ans;
    }
};
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