675. Cut Off Trees for Golf Event

Question

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 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

• Method 1: PriorityQueue, bfs
  • The idea of this method is we sort all nodes with their values.
  • Then we use the order to find the minimum path.

      class Solution {        
      public int cutOffTree(List<List<Integer>> forest) {
          int height = forest.size(), width = forest.get(0).size();
          int[][] map = new int[height][width];
          PriorityQueue<int[]> pq = new PriorityQueue(new Comparator<int[]>(){
              @Override
              public int compare(int[] n1, int[] n2){
                  return map[n1[0]][n1[1]] - map[n2[0]][n2[1]];
              }
          });
          for(int i = 0; i < height; i++){
              for(int j = 0; j < width; j++){
                  map[i][j] = forest.get(i).get(j);
                  if(map[i][j] != 0 && map[i][j] != 1)  pq.offer(new int[]{i, j});
              }
          }
          if(map[0][0] == 0) return -1;
          int res = 0;
          Queue<int[]> queue = new LinkedList<>();
          int[] cur = new int[]{0, 0};
          while(!pq.isEmpty()){
              int[] dest = pq.poll();
              queue.clear();
              queue.offer(cur);
              boolean found = false;
              boolean[][] visited = new boolean[height][width];
              visited[cur[0]][cur[1]] = true;
              int step = -1;
              while(!queue.isEmpty() && !found){
                  ++step;
                  int size = queue.size();
                  for(int i = 0; i < size; i++){
                      int[] node = queue.poll();
                      if(map[node[0]][node[1]] == map[dest[0]][dest[1]]){
                          found = true;
                          res += step;
                          break;
                      }
                      int tempRow = 0, tempCol = 0;
                      for(int d = 0; d < 4; d++){
                          tempRow = node[0] + dir[d][0];
                          tempCol = node[1] + dir[d][1];
                          if(tempRow >= 0 && tempRow < height && tempCol >= 0 && tempCol < width && map[tempRow][tempCol] != 0 && (map[tempRow][tempCol] >= map[node[0]][node[1]] || !visited[tempRow][tempCol])){
                              visited[tempRow][tempCol] = true;
                              queue.add(new int[]{tempRow, tempCol});
                          }
                      }
                  }
                  if(found) break;
              }
              if(found) cur = dest;
              else return -1;
          }
          return res;
      }
      }
    

Reference

1. 花花酱 LeetCode 675. Cut Off Trees for Golf Event - 刷题找工作 EP55