<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 {
    int n, m;
    vector<vector<bool>> vis;
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {-1, 1, 0, 0};
    bool isvalid(int x, int y) {
        return x >=0 and y >= 0 and x < n and y < m;
    } 

    int shortestdist(int stx, int sty, int enx, int eny, vector<vector<int>>& forest) {
        fill(vis.begin(), vis.end(), vector<bool>(m, 0));
        queue<pair<int, pair<int,int>>> q;
        q.push({0, {stx, sty}});
        while(!q.empty()) {
            int d = q.front().first;
            auto [x, y] = q.front().second;
            q.pop();

            if (x == enx and y == eny) return d;
            if (vis[x][y]) continue;
            vis[x][y] = 1;

            for (int k=0; k<4; k++) {
                int i = x + dx[k];
                int j = y + dy[k];
                if (!isvalid(i, j) || !forest[i][j]) continue;
                q.push({d+1, {i, j}});
            }
        }
        return -1;
    }

public:
    int cutOffTree(vector<vector<int>>& forest) {
        n = forest.size();
        m = forest.back().size();
        vis.resize(n, vector<bool>(m));

        map<int, pair<int,int>> mp;
        for (int i=0; i<n; i++) 
            for (int j=0; j<m; j++) 
                if (forest[i][j] > 1) 
                    mp.insert({forest[i][j], {i,j}});

        pair<int, int> curr = {0, 0};
        int dist = 0;
        for (auto it: mp) {
            auto [x, y] = it.second;
            auto [cx, cy] = curr;
            int incr = shortestdist(x, y, cx, cy, forest);
            if (incr < 0) return -1;
            dist += incr; 
            curr = {x, y};
        }
        return dist;
    }
};
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