# Cut Off Trees for Golf Event

`Graph`, `Breadth-first Search`, `Shortest Path`

Hard

You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map, in this map:

1. `0` represents the `obstacle` can't be reached.

2. `1` represents the `ground` can be walked through.

3. `The place with number bigger than 1` represents a `tree` can be walked through, and this positive number represents the tree's height.

You are asked to cut off all the trees in this forest in the order of tree's height - always cut off the tree with lowest height first. And after cutting, the original place has the tree will become a grass (value 1).

You will start from the point (0, 0) and you should output the minimum steps you need to walk to cut off all the trees. If you can't cut off all the trees, output -1 in that situation.

You are guaranteed that no two`trees`have the same height and there is at least one tree needs to be cut off.

Example 1:

``````Input:

[
[1,2,3],
[0,0,4],
[7,6,5]
]

Output:
6``````

Example 2:

``````Input:

[
[1,2,3],
[0,0,0],
[7,6,5]
]

Output:
-1``````

Example 3:

``````Input:

[
[2,3,4],
[0,0,5],
[8,7,6]
]

Output:
6

Explanation:
You started from the point (0,0) and you can cut off the tree in (0,0) directly without walking.``````

Hint: size of the given matrix will not exceed 50x50.

## Solution & Analysis

### 基本框架：

• 先找到所有的树，按照高度从低到高排序

• 从(0,0)作为起点，每次都寻找到下一棵树的最短路径长度，并更新总步数长度；以及更新起点位置和下一个目标位置

• 如果最短路径不存在，返回-1。

### BFS

``````class Solution {
public int cutOffTree(List<List<Integer>> forest) {
// Get all the trees with h-height, r-row, c-column and sort by height
List<int[]> trees = new ArrayList<>();
int[][] map = new int[forest.size()][forest.get(0).size()];
for (int r = 0; r < forest.size(); r++) {
for (int c = 0; c < forest.get(r).size(); c++) {
int h = forest.get(r).get(c);
map[r][c] = h;
if (h > 1) {
}
}
}

// Sort trees based on height of trees
Collections.sort(trees, (t1, t2) -> Integer.compare(t1[0], t2[0]));

int ans = 0;
int r0 = 0;
int c0 = 0;
// Find distance from current position to next tree
for (int[] tree: trees) {
int d = dist(map, r0, c0, tree[1], tree[2]);
if (d < 0) {
return -1;
}
ans += d;
r0 = tree[1];
c0 = tree[2];
}
return ans;
}

private static int[][] dirs = {{-1, 0}, {1, 0} , {0, -1}, {0, 1}};

private int dist(int[][] map, int sr, int sc, int tr, int tc) {
int distance = 0;
boolean[][] visited = new boolean[map.length][map[0].length];
q.offer(new int[] {sr, sc});
visited[sr][sc] = true;

while (!q.isEmpty()) {
int levelSize = q.size();
while (levelSize-- > 0) {
int[] pos = q.poll();

if (pos[0] == tr && pos[1] == tc) {
return distance;
}

for (int[] dir: dirs) {
int nr = pos[0] + dir[0];
int nc = pos[1] + dir[1];
if (nr >= 0 && nr < map.length && nc >= 0 && nc < map[0].length) {
if (map[nr][nc] != 0 && !visited[nr][nc]) {
q.offer(new int[] {nr, nc});
visited[nr][nc] = true;
}
}
}
}
distance++;
}
return -1;
}
}``````

### Another BFS (Store distance in int[] {r, c, dist})

Runtime: 258 ms, faster than 84.21% of Java online submissions for Cut Off Trees for Golf Event.

Memory Usage: 46.7 MB, less than 100.00% of Java online submissions for Cut Off Trees for Golf Event.

``````class Solution {
public int cutOffTree(List<List<Integer>> forest) {
// Get all the trees with h-height, r-row, c-column and sort by height
List<int[]> trees = new ArrayList<>();
int[][] map = new int[forest.size()][forest.get(0).size()];
for (int r = 0; r < forest.size(); r++) {
for (int c = 0; c < forest.get(r).size(); c++) {
int h = forest.get(r).get(c);
map[r][c] = h;
if (h > 1) {
}
}
}

// Sort trees based on height of trees
Collections.sort(trees, (t1, t2) -> Integer.compare(t1[0], t2[0]));

int ans = 0;
int r0 = 0;
int c0 = 0;
// Find distance from current position to next tree
for (int[] tree: trees) {
int d = dist(forest, r0, c0, tree[1], tree[2]);
if (d < 0) {
return -1;
}
ans += d;
r0 = tree[1];
c0 = tree[2];
}
return ans;
}

private static int[][] dirs = {{-1, 0}, {1, 0} , {0, -1}, {0, 1}};

public int dist(List<List<Integer>> forest, int sr, int sc, int tr, int tc) {
int R = forest.size(), C = forest.get(0).size();
boolean[][] seen = new boolean[R][C];

seen[sr][sc] = true;

while (!queue.isEmpty()) {
int[] cur = queue.poll();
if (cur[0] == tr && cur[1] == tc) {
return cur[2];
}
for (int[] dir: dirs) {
int r = cur[0] + dir[0];
int c = cur[1] + dir[1];
if (0 <= r && r < R && 0 <= c && c < C &&
!seen[r][c] && forest.get(r).get(c) > 0) {
seen[r][c] = true;
queue.add(new int[]{r, c, cur[2] + 1});
}
}
}
return -1;
}
}``````

### Dijkstra's Algorithm (Greedy + PriorityQueue based)

BFS区别就在于不再是一层一层遍历，而是每次通过在PriorityQueue`poll()`，得到当前待遍历的节点中到起始点距离最短的节点，然后四方向寻找，如果发现距离比`distance[][]` matrix所记录的要小，就更新`distance[][]`matrix，并且将该节点以及其到起始点的距离放入PriorityQueue中，进行下一轮`poll()`的过程。

Performance:

• Runtime: 911 ms, faster than 3.64% of Java online submissions for Cut Off Trees for Golf Event.

• Memory Usage: 46.7 MB, less than 100.00% of Java online submissions for Cut Off Trees for Golf Event.

``````class Solution {
public int cutOffTree(List<List<Integer>> forest) {
// Get all the trees with h-height, r-row, c-column and sort by height
List<int[]> trees = new ArrayList<>();
int[][] map = new int[forest.size()][forest.get(0).size()];
for (int r = 0; r < forest.size(); r++) {
for (int c = 0; c < forest.get(r).size(); c++) {
int h = forest.get(r).get(c);
map[r][c] = h;
if (h > 1) {
}
}
}

// Sort trees based on height of trees
Collections.sort(trees, (t1, t2) -> Integer.compare(t1[0], t2[0]));

int ans = 0;
int r0 = 0;
int c0 = 0;
// Find distance from current position to next tree
for (int[] tree: trees) {
int d = dist(map, r0, c0, tree[1], tree[2]);
if (d < 0) {
return -1;
}
ans += d;
r0 = tree[1];
c0 = tree[2];
}
return ans;
}

private static int[][] dirs = {{-1, 0}, {1, 0} , {0, -1}, {0, 1}};

// Dijkstra's Algorithm to Find Shortest Distance
private int dist(int[][] map, int sr, int sc, int tr, int tc) {

// Init distance matrix
int[][] distance = new int[map.length][map[0].length];
for (int[] row: distance) {
Arrays.fill(row, Integer.MAX_VALUE);
}
distance[sr][sc] = 0;

// Init priority queue for vertices to traverse
PriorityQueue<int[]> q = new PriorityQueue<int[]>((a, b) -> a[2] - b[2]);
q.offer(new int[] {sr, sc, 0});

while (!q.isEmpty()) {
int[] pos = q.poll();

for (int[] dir: dirs) {
int nr = pos[0] + dir[0];
int nc = pos[1] + dir[1];
if (nr >= 0 && nr < map.length && nc >= 0 && nc < map[0].length) {
// Update distance matrix and add the vertex to priority queue
if (map[nr][nc] != 0 && (pos[2] + 1 < distance[nr][nc])) {
distance[nr][nc] = pos[2] + 1;
q.offer(new int[] {nr, nc, pos[2] + 1});
}
}
}
}

if (distance[tr][tc] == Integer.MAX_VALUE) {
return -1;
}
return distance[tr][tc];
}
}``````