<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:
  int n, m;
  vector<vector<int>> forest;
  vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

  bool inGrid(pair<int, int> u) {
    return u.first >= 0 && u.first < n && u.second >= 0 && u.second < m;
  }

  int bfs(pair<int, int> source, pair<int, int> target) {
    queue<pair<int, int>> q;
    vector<vector<int>> vis(n, vector<int>(m, 0));
    vector<vector<int>> dist(n, vector<int>(m, 0));

    q.push(source);
    vis[source.first][source.second] = 1;

    while (!q.empty()) {
      auto u = q.front();
      q.pop();

      if (u == target) {
        return dist[u.first][u.second];
      }

      for (auto [d1, d2] : dir) {
        pair<int, int> v = {u.first + d1, u.second + d2};

        if (!inGrid(v) || vis[v.first][v.second] || !forest[v.first][v.second])
          continue;

        vis[v.first][v.second] = 1;
        dist[v.first][v.second] = dist[u.first][u.second] + 1;
        q.push(v);
      }
    }

    return -1;
  }

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

    this->forest = forest;

    vector<tuple<int, int, 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});
        }
      }
    }

    pair<int, int> start = {0, 0};
    sort(begin(positions), end(positions));

    for (auto [val, i, j] : positions) {
      int steps = bfs({start.first, start.second}, {i, j});

      if (steps == -1) {
        return -1;
      }

      res += steps;
      start = {i, j};
    }

    return res;
  }
};
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