<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 r=forest.size(), c=forest[0].size();
        vector<pair<int, pair<int, int>>>v;
        int maxi=-1;
        for(int i=0;i<r;i++){
            for(int j=0;j<c;j++){
                if(forest[i][j]>1){
                    v.push_back({forest[i][j], {i, j}});
                }
            }
        }
        int ans=0;
        int x=0, y=0;
        sort(v.begin(), v.end());
        int n=v.size();
        for(int i=0;i<n;i++){
            int dx=v[i].second.first;
            int dy=v[i].second.second;
            int val=dijkstra(forest, x, y, dx, dy);
            x=dx, y=dy;
            if(val==-1) return -1;
            ans+=val;
        }
        return ans;
    }
    int dijkstra(vector<vector<int>>&f, int x, int y, int dx, int dy){
        int r=f.size();
        int c=f[0].size();
        queue<pair<int, pair<int, int>>>pq;
        vector<vector<int>>vis(r, vector<int>(c, -1));
        pq.push({0, {x, y}});
        int nr[]={0, -1, 0, 1};
        int nc[]={1, 0, -1, 0};
        while(pq.empty()==false){
            int steps=pq.front().first;
            int i=pq.front().second.first;
            int j=pq.front().second.second;
            pq.pop();
            if(vis[i][j]==1)continue;
            vis[i][j]=1;
            if(i==dx&&j==dy) return steps;
            for(int k=0;k<4;k++){
                int ni=nr[k]+i;
                int nj=nc[k]+j;
                if(ni>=0&&ni<r&&nj>=0&&nj<c&&f[ni][nj]>0&&vis[ni][nj]==-1){
                    pq.push({steps+1, {ni, nj}});
                }
            }
        }
        return -1;
    }
};
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