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

    bool valid(int x,int y,int m,int n){
        if(x<0||y<0||x>=m||y>=n)return 0;
        return 1;
    }

    int rec(int ss,int sd,int es,int ed,vector<vector<int>>& forest){
        if(ss==es&&sd==ed)return 0;
        int m=forest.size();
        int n=forest[0].size();
        vector<vector<int>> dist(m,vector<int>(n,1e9));
        vector<vector<int>> vis(m,vector<int>(n,0));
        queue<pair<int,int>> q;
        q.push({ss,sd});
        dist[ss][sd]=0;
        while(!q.empty()){
            int x=q.front().first;
            int y=q.front().second;
            q.pop();
            if(vis[x][y])continue;
            vis[x][y]=1;
            for(int k=0;k<4;++k){
                int a=x+dx[k];
                int b=y+dy[k];
                if((valid(a,b,m,n))&&(forest[a][b]!=0)&&(dist[a][b]>dist[x][y]+1)){
                    dist[a][b]=dist[x][y]+1;
                    q.push({a,b});
                }
            }
        }
        if(dist[es][ed]==1e9)return -1;
        return dist[es][ed];
    }

    int cutOffTree(vector<vector<int>>& forest) {
        int m=forest.size();
        int n=forest[0].size();
        vector<vector<int>> tree;
        for(int i=0;i<m;++i){
            for(int j=0;j<n;++j){
                if(forest[i][j]>1){
                    tree.push_back({forest[i][j],i,j});
                }
            }
        }
        sort(tree.begin(),tree.end());
        int x=0;
        int y=0;
        int ans=0;
        for(int i=0;i<tree.size();++i){
            int step=rec(x,y,tree[i][1],tree[i][2],forest);
            if(step==-1)return -1;
            ans+=step;
            x=tree[i][1];
            y=tree[i][2];
        }
        return ans;
    }
};
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