<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>>& f) {
        // shortest distance to next largest such that there is no obstacle in between
        constexpr int inf = 1E5;
        constexpr int dx[] = {-1, 1, 0, 0};
        constexpr int dy[] = {0, 0, -1, 1};
        const int m = f.size();
        const int n = f[0].size();

        auto minDis = [&] (int a, int b, int x, int y) -> int {
            // minimum distance from (a, b) to (x, y)
            queue<pair<int, int>> q;
            vector dist(m, vector<int> (n, -1));

            q.push({a, b});
            dist[a][b] = 0;

            while (!q.empty()) {
                auto z = q.front();
                q.pop();
                
                for (int i = 0; i < 4; i++) {
                    int xx = z.first + dx[i];
                    int yy = z.second + dy[i];

                    if (xx < 0 || xx >= m || yy < 0 || yy >= n || dist[xx][yy] != -1 || f[xx][yy] == 0) {
                        continue;
                    }

                    q.push({xx, yy});
                    dist[xx][yy] = 1 + dist[z.first][z.second];
                }
            } 

            if (dist[x][y] == -1) {
                return -inf;
            }
            return dist[x][y];
        };

        map<int, pair<int, int>> mp;
        vector<int> nums;
        for (int i = 0; i < (int) f.size(); i++) {
            for (int j = 0; j < (int) f[i].size(); j++) {
                if (f[i][j] > 1) {
                    nums.push_back(f[i][j]);
                    mp[f[i][j]] = {i, j};
                }
            }
        }

        sort(nums.begin(), nums.end());
        vector<pair<int, int>> toGo;
        for (int i = 0; i < nums.size(); i++) {
            toGo.push_back(mp[nums[i]]);
        }

        int ans = 0;
        pair<int, int> start = {0, 0};
        for (int i = 0; i < toGo.size(); i++) {
            auto end = toGo[i];
            ans += minDis(start.first, start.second, end.first, end.second);
            start = end;
        }

        if (ans < 0) {
            ans = -1;
        }

        return ans;
    }
};
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