<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 xx[4] = {0, 0, 1, -1};
    int yy[4] = {1, -1, 0, 0};
    int minDis(vector<vector<int>>& forest, pair<int,int> a, pair<int,int> b){
        int n = forest.size();
        int m = forest[0].size();
        queue<pair<pair<int,int>,int>> q;
        q.push({{a.first, a.second}, 0});
        vector<vector<int>> vis(n, vector<int>(m));
        while(q.size()){
            auto top= q.front();
            q.pop();
            int i = top.first.first;
            int j = top.first.second;
            int dis = top.second;
            if(top.first==b) return dis;
            if(vis[i][j]) continue;
            vis[i][j] = 1;
            for(int k=0;k<4;k++){
                int x = i+xx[k];
                int y = j+yy[k];
                if(x<0 || x>=n || y<0 || y>=m || vis[x][y] || !forest[x][y]) continue;
                q.push({{x, y}, dis+1});
            }
        }
        return -1;
    }
    int cutOffTree(vector<vector<int>>& forest) {
        vector<pair<int,pair<int,int>>> values;
        int n = forest.size();
        int m = forest[0].size();

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(forest[i][j]>1){
                    values.push_back({forest[i][j], {i, j}});
                }
            }
        }
        sort(values.begin(), values.end(), greater<pair<int,pair<int,int>>>());
        values.push_back({1, {0, 0}});
        int ans = 0;
        for(int i=values.size()-2;i>=0;i--){
            int curDis = minDis(forest, values[i].second, values[i+1].second);
            if(curDis==-1) return -1;
            ans+=curDis;
        }
        return ans;
    }
};
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