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)
use std::collections::HashSet;
use std::collections::VecDeque;

const DIRS: [[i32; 2]; 4] = [[-1, 0], [1, 0], [0, -1], [0, 1]];

impl Solution {
    pub fn cut_off_tree(forest: Vec<Vec<i32>>) -> i32 {
        let (row, col) = (forest.len() as i32, forest[0].len() as i32);

        let bfs = |start: i32, end: i32| -> i32 {
            let mut queue = VecDeque::new();
            let mut vis = HashSet::new();
            queue.push_back(start);
            vis.insert(start);
            let mut step = 0;
            while !queue.is_empty() {
                let n = queue.len();
                for _ in 0..n {
                    let state = queue.pop_front().unwrap();
                    if state == end {
                        return step;
                    }
                    for k in 0..4 {
                        let x = state / col + DIRS[k][0];
                        let y = (state % col) + DIRS[k][1];
                        let nxt = x * col + y;
                        if x >= 0
                            && x < row
                            && y >= 0
                            && y < col
                            && forest[x as usize][y as usize] != 0
                            && !vis.contains(&nxt)
                        {
                            queue.push_back(nxt);
                            vis.insert(nxt);
                        }
                    }
                }
                step += 1;
            }
            -1
        };

        let mut trees = Vec::new();
        for i in 0..row {
            for j in 0..col {
                let height = forest[i as usize][j as usize];
                if height > 1 {
                    trees.push((height, i * col + j));
                }
            }
        }
        trees.sort();

        let (mut ans, mut start) = (0, 0);
        for t in &trees {
            let end = t.1;
            let step = bfs(start, end);
            if step == -1 {
                return -1;
            }
            ans += step;
            start = end;
        }
        ans
    }
}
