Cut Off Trees for Golf Event
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0675.Cut%20Off%20Trees%20for%20Golf%20Event/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0675.Cut%20Off%20Trees%20for%20Golf%20Event/images/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>
Breadth-First Search; Array; Matrix; Heap (Priority Queue)
class Solution {
public:
    int m;
    int n;
    vector<int> dist;

    int cutOffTree(vector<vector<int>>& forest) {
        m = forest.size();
        n = forest[0].size();
        dist.resize(3600);
        vector<pair<int, int>> trees;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (forest[i][j] > 1)
                    trees.push_back({forest[i][j], i * n + j});
        sort(trees.begin(), trees.end());
        int ans = 0;
        int start = 0;
        for (auto& tree : trees) {
            int end = tree.second;
            int t = bfs(start, end, forest);
            if (t == -1) return -1;
            ans += t;
            start = end;
        }
        return ans;
    }

    int bfs(int start, int end, vector<vector<int>>& forest) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
        q.push({f(start, end), start});
        fill(dist.begin(), dist.end(), INT_MAX);
        dist[start] = 0;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        while (!q.empty()) {
            int state = q.top().second;
            q.pop();
            if (state == end) return dist[state];
            for (int k = 0; k < 4; ++k) {
                int x = state / n + dirs[k], y = state % n + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && forest[x][y] && dist[x * n + y] > dist[state] + 1) {
                    dist[x * n + y] = dist[state] + 1;
                    q.push({dist[x * n + y] + f(x * n + y, end), x * n + y});
                }
            }
        }
        return -1;
    }

    int f(int start, int end) {
        int a = start / n, b = start % n;
        int c = end / n, d = end % n;
        return abs(a - c) + abs(b - d);
    }
};