<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:
    bool valid(int x, int y, int m, int n) {
        return (x >= 0 && x < m && y >= 0 && y < n);
    }
    int nextRow[4] = {0, 0, 1, -1};
    int nextCol[4] = {1, -1, 0, 0};
    int bfs(vector<vector<int>> forest, int x, int y, int nextx, int nexty) {
        int m = forest.size();
        int n = forest[0].size();

        queue<pair<int, pair<int, int>>>q;

        vector<vector<int>> visited(m, vector<int>(n, 0));

        visited[x][y] = 1;

        q.push({0, {x, y}});
        while(!q.empty()) {

            int row = q.front().second.first;
            int col = q.front().second.second;
            
            int dist = q.front().first;
            
            q.pop();
            if(row == nextx && col == nexty) return dist;
            for(int i = 0; i < 4 ; i++) {
                int r = row + nextRow[i];
                int c = col + nextCol[i];
                if(valid(r, c, m, n) && !visited[r][c] && forest[r][c]) {
                    visited[r][c] = true;
                    q.push({dist + 1, {r, c}}) ;
                }
            }
        }

        return -1;
    }
    int cutOffTree(vector<vector<int>>& forest) {
        int m = forest.size();
        int n = forest[0].size();
        vector<pair<int, pair<int, int>>>temp;
        for(int i = 0; i < m ; i++) {
            for(int j = 0; j < n ; j++) {
                if(forest[i][j] > 1) {
                    temp.push_back({forest[i][j], {i, j}});
                }
            }
        }
        temp.push_back({0, {0, 0}});
        sort(temp.begin(), temp.end());
        int ans = 0;
        for(int i = 0 ; i < temp.size() - 1; i++) {
            int dist = bfs(forest, temp[i].second.first, temp[i].second.second, temp[i+1].second.first, temp[i+1].second.second);

            if(dist == -1) return -1;

            ans += dist;
        }
        return ans;
    }
};
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