<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 shortestPath(pair<int, int> start, pair<int, int> end, vector<vector<int>>& forest){
        queue<pair<pair<int, int>, int>> q;
        q.push({{start.first, start.second}, 0});
        int rows = forest.size();
        int cols = forest[0].size();
        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        while(!q.empty()) {
            auto curr = q.front();
            auto pos = curr.first;
            q.pop();
            if(visited[pos.first][pos.second])
                continue;
            else
                visited[pos.first][pos.second] = true;
            if(pos == end)
                return curr.second;
            else{
                if(pos.first > 0 && forest[pos.first - 1][pos.second])
                    q.push({{pos.first - 1, pos.second}, curr.second + 1});
                if(pos.second > 0 && forest[pos.first][pos.second - 1])
                    q.push({{pos.first, pos.second - 1}, curr.second + 1});
                if(pos.first < forest.size() - 1 && forest[pos.first + 1][pos.second])
                    q.push({{pos.first + 1, pos.second}, curr.second + 1});
                if(pos.second < forest[0].size() - 1 && forest[pos.first][pos.second + 1])
                    q.push({{pos.first, pos.second + 1}, curr.second + 1});
            }
        }
        return -1;
    }

    int cutOffTree(vector<vector<int>>& forest) {
        if(!forest[0][0])
            return -1;
        map<int, pair<int, int>> lookup;
        for(int i=0;i<forest.size();i++)
        {
            for(int j=0;j<forest[0].size();j++)
                lookup[forest[i][j]] = {i, j};
        }
        lookup.erase(0);
        lookup.erase(1);
        pair<int, int> currentPos = {0, 0};
        int result = 0;
        for(auto& v: lookup)
        {
            pair<int, int> targetPos = v.second;
            int pathLength = shortestPath(currentPos, targetPos, forest);
            if(pathLength < 0)
                return -1;
            else
                result += pathLength;
            currentPos = targetPos;
        }
        return result;      
    }
};
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