<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:
    vector<pair<int,int>> moves = {{0,1},{0,-1}, {1,0}, {-1, 0}};
    int cutOffTree(vector<vector<int>>& forest) {
        int n = forest.size(), m = forest[0].size();
        vector<pair<int,int>> vp;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                int coord = i*n*m + j;
                if (forest[i][j] > 1) {
                    vp.push_back({forest[i][j], coord});
                }
            }
        }
        sort(vp.begin(), vp.end());
        int src = 0, ans = 0;
        for(auto [v,coord]: vp) {
            int len = shortestPath(src, coord, forest, n, m);
            if (len == -1) {
                return -1;
            }
            src = coord;
            ans += len;
        }
        return ans;
    }

    int shortestPath(int src, int dest, vector<vector<int>>& forest, int n, int m) {
        if (src == dest) {
            return 0;
        }
        vector<vector<bool>> vis(n, vector<bool>(m, false));
        queue<pair<int,int>> q;
        q.push({src, 0});
        vis[src/(n*m)][src%(n*m)] = true;
        while(!q.empty()) {
            auto f = q.front();
            int i = f.first/(m*n);
            int j = f.first%(m*n);
            int dis = f.second;
            q.pop();
            for(auto mv: moves) {
                int a = mv.first + i;
                int b = mv.second + j;
                if (a >= 0 && a < n && b >= 0 && b < m && forest[a][b] != 0) {
                    int coord = (a*(m*n) + b);
                    if (!vis[a][b]) {
                        if (dest == coord) {
                            return dis + 1;
                        }
                        q.push({coord, dis+1});
                        vis[a][b] = true;
                    }
                }
            }
        }
        return -1;
    }
};
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