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:

0 represents the obstacle can't be reached.
1 represents the ground can be walked through.
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 twotreeshave 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

跟The Maze系列有些类似；都是寻找最短路径，但是这里要计算每次从新起点到下一棵要砍的树的最短距离，所以shortest distance需要被call多次（每次砍树都要先call）。

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。

最短路径的寻找方式有许多种，在grid类型的graph中，有十分简便的实现：

因为grid类型的graph可以看成edge的weight/length均为1的无向图，因此BFS可以一层一层地遍历，到达目标点的总步长也就是遍历的层数，每一层开始前记录当前层中节点数量，遍历完之后，再遍历下一层，放在queue中的是每一层的节点。如果找到目标位置，则返回当前的层数即可。

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) {
                    trees.add(new int[] {h, r, c});
                }
            }
        }

        // 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];
        Queue<int[]> q = new LinkedList<int[]>();
        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})

类似上面的一层一层BFS，这里就是把距离放到每个节点的int[]中.

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) {
                    trees.add(new int[] {h, r, c});
                }
            }
        }

        // 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();
        Queue<int[]> queue = new LinkedList();
        boolean[][] seen = new boolean[R][C];

        queue.add(new int[]{sr, sc, 0});
        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)

与BFS的区别就在于不再是一层一层遍历，而是每次通过在PriorityQueue中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) {
                    trees.add(new int[] {h, r, c});
                }
            }
        }

        // 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];
    }
}

Approach: A* Search

TODO: https://leetcode.com/articles/cutoff-trees-for-golf-event/?orderBy=most_votes

Approach: Hadlock's Algorithm

TODO: https://leetcode.com/articles/cutoff-trees-for-golf-event/?orderBy=most_votes

Reference

http://www.cnblogs.com/grandyang/p/8379506.html

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