<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 dist = 0;

  int dx[4]{-1, 0, 0, 1};
  int dy[4]{0, -1, 1, 0};

  vector<pair<int, pair<int, int>>> points;
  for (int i = 0; i < forest.size(); i++) {
    for (int j = 0; j < forest[0].size(); j++) {
      if (forest[i][j] > 1)
        points.push_back({forest[i][j], {i, j}});
    }
  }
  if (points.size() == 0)
    return 0;

  sort(points.begin(), points.end());

  for (int i = 0; i < points.size(); i++) {
    int targetI = points[i].second.first;
    int targetJ = points[i].second.second;
    vector<vector<bool>> visited(forest.size(), vector<bool>(forest[0].size()));
    queue<pair<int, pair<int, int>>> q;
    if (i > 0) {
      q.push({0, {points[i-1].second.first, points[i-1].second.second}});
    } else {
      q.push({0, {0, 0}});
    }

    while (!q.empty()) {
      int curDist = q.front().first;
      int curI = q.front().second.first;
      int curJ = q.front().second.second;
      q.pop();

      if (visited[curI][curJ])
        continue;
      visited[curI][curJ] = true;

      if (curI == targetI && curJ == targetJ) {
        dist += curDist;
        forest[curI][curJ] = 1;
        break;
      }

      for (int k = 0; k < 4; k++) {
        int nI = curI + dx[k];
        int nJ = curJ + dy[k];
        if (nI >= 0 && nI < forest.size() && nJ >= 0 && nJ < forest[0].size() &&
            !visited[nI][nJ] && forest[nI][nJ] > 0) {
          q.push({curDist + 1, {nI, nJ}});
        }
      }
    }

    if (!visited[targetI][targetJ]) {
      return -1;
    }
  }

  return dist;
}


};
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