<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<vector<int>> adj(m * n);
        vector<pair<int, int>> list;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(forest[i][j] != 0){
                    if(forest[i][j] != 1){
                        list.push_back({forest[i][j], i * n + j});
                    }
                    if(i + 1 < m && forest[i + 1][j] != 0){
                        adj[i * n + j].push_back((i + 1) * n + j);
                        adj[(i + 1) * n + j].push_back(i * n + j);
                    }
                    if(j + 1 < n && forest[i][j + 1] != 0){
                        adj[i * n + j].push_back(i * n + j + 1);
                        adj[i * n + j + 1].push_back(i * n + j);
                    }
                }
            }
        }
        sort(list.begin(), list.end());
        int ans = 0, prev = 0;
        for(int i = 0; i < list.size(); i++){
            queue<int> q;
            vector<int> d(m * n, 0);
            vector<bool> used(m * n, false);
            q.push(prev);
            d[prev] = 0;
            used[prev] = true;
            if(prev == list[i].second){
                continue;
            }
            while(!q.empty()){
                int v = q.front();
                q.pop();
                for(int u: adj[v]){
                    if(!used[u]){
                        used[u] = true;
                        d[u] = d[v] + 1;
                        q.push(u);
                        if(u == list[i].second){
                            ans += d[u];
                            break;
                        }
                    }
                }
                if(used[list[i].second]){
                    break;
                }
            }
            if(!used[list[i].second]){
                return -1;
            }
            prev = list[i].second;
        }
        return ans;
    }
};
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