<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>

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    int n, m;
    int bfs(int si, int sj, int ti, int tj, vector<vector<int>> &forest)
    {
        vector<vector<int>> vis(n, vector<int>(m, 0));
        vis[si][sj] = 1;
        vector<vector<int>> dist(n, vector<int>(m, 0));
        // dist[si][sj] = 0;
        queue<pair<int, int>> q;
        q.push({si,sj});
        while (!q.empty())
        {
            auto p = q.front();
            q.pop();
            int i = p.first, j = p.second;
            if (i == ti && j == tj)
                return dist[i][j];
            for (auto d : dir)
            {
                int ni = i + d.first, nj = j + d.second;
                if (ni >= 0 && ni < n && nj >= 0 && nj < m && forest[ni][nj] != 0 && vis[ni][nj] == 0)
                {
                    vis[ni][nj] = 1;
                    dist[ni][nj] = dist[i][j] + 1;
                    q.push({ni, nj});
                }
            }
        }
        return -1;
    }
    int cutOffTree(vector<vector<int>> &forest)
    {
        n = forest.size(), m = forest[0].size();
        vector<vector<int>> positions;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (forest[i][j] > 1)
                    positions.push_back({forest[i][j], i, j});
            }
        }
        sort(positions.begin(), positions.end());
        int si = 0, sj = 0;
        int ans = 0;
        for (auto p : positions)
        {
            int ti = p[1], tj = p[2];
            int d = bfs(si, sj, ti, tj, forest);
            if (d == -1)
                return -1;
            ans += d;
            // forest[ti][tj] = 1;
            si = ti, sj = tj;
        }
        return ans;
    }
};
384465