<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>

// bfs
#define INF 1e9
class Solution {
public:
    int cutOffTree(vector<vector<int>>& a) {
        int n=a.size(),m=a[0].size();
        vector<pair<int,pair<int,int>>> v;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if (a[i][j]>1) v.push_back({a[i][j],{i,j}}); // collect cac vi tri co cay
            }
        }
        sort(v.begin(),v.end()); // sap xep theo chieu cao cay tang dan
       
        int i=0,sum=0, is=0,js=0; // bat dau tu [0][0]
        while(i<v.size()){
            auto x=v[i].second;
            int d=helper(is,js, x.first,x.second,a);
            if (d==INF) return -1;
            sum+=d;
            is=x.first;
            js=x.second;
            i++;
        }
        return sum;
    }
    int helper(int is, int js, int id, int jd, vector<vector<int>>& a){
        int dx[]={0,0,-1,1};
        int dy[]={1,-1,0,0};
        int n=a.size(),m=a[0].size();
        queue<tuple<int,int,int>> q;//i,j, distance // chuyen tu vector qua mang tinh hoac tuple-> giam time nhieu, (queue<vector<int>> q)
        // https://stackoverflow.com/questions/15166879/queue-containing-fixed-size-arrays-in-c
        q.push({is,js,0});
        vector<vector<bool>> v(n,vector<bool>(m));
        v[is][js]=true;
        while(!q.empty()){
            auto cur=q.front();
            q.pop();
            int i=get<0>(cur),j=get<1>(cur),d=get<2>(cur);
            if (i==id && j==jd) return d;
            for(int t=0;t<4;t++){
                int ni=i+dx[t],nj=j+dy[t];
                if (ni<0 || ni>=n || nj<0 || nj>=m || v[ni][nj] || a[ni][nj]==0) continue;
                if (ni==id && nj==jd) return d+1;// thoat truoc de tranh TLE -> nhung ko an thua lam nhe
                q.push({ni,nj,d+1});
                v[ni][nj]=true;
            }
        }
        return INF;
    }
};
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