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This approach involves using a BFS algorithm to find the shortest path between the starting point and each tree in increasing order of their heights. We first list all trees with their positions and heights, sort them, and for each sorted tree, compute the shortest path using BFS. If any tree is unreachable, we return -1.
The time complexity of this approach is O(T * m * n) where T is the number of trees. This is because we perform BFS from the starting position to each tree. The space complexity is O(m * n) due to the storage of the queue and visited matrix in BFS.
1#include <stdio.h>
2#include <stdlib.h>
3#include <stdbool.h>
4#include <limits.h>
5
6// Define directions for moving in the grid
7int directions[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
8
9// Function prototypes
10int bfs(int** forest, int m, int n, int sr, int sc, int tr, int tc);
11int cmpfunc(const void* a, const void* b);
12
13int cutOffTree(int** forest, int forestSize, int* forestColSize){
14 int m = forestSize;
15 int n = forestColSize[0];
16
17 // Extract trees and sort by height
18 int trees[m*n][3], count = 0;
19 for (int i = 0; i < m; i++) {
20 for (int j = 0; j < n; j++) {
21 if (forest[i][j] > 1) {
22 trees[count][0] = forest[i][j];
23 trees[count][1] = i;
24 trees[count][2] = j;
25 count++;
26 }
27 }
28 }
29 qsort(trees, count, sizeof(trees[0]), cmpfunc);
30
31 // Start cutting trees
32 int totalSteps = 0, sr = 0, sc = 0;
33 for (int i = 0; i < count; i++) {
34 int tr = trees[i][1];
35 int tc = trees[i][2];
36 int steps = bfs(forest, m, n, sr, sc, tr, tc);
37 if (steps == -1) return -1;
38 totalSteps += steps;
39 sr = tr;
40 sc = tc;
41 }
42 return totalSteps;
43}
44
45int bfs(int** forest, int m, int n, int sr, int sc, int tr, int tc) {
46 if (sr == tr && sc == tc) return 0;
47 bool visited[m][n];
48 for (int i = 0; i < m; i++) {
49 for (int j = 0; j < n; j++) {
50 visited[i][j] = false;
51 }
52 }
53 visited[sr][sc] = true;
54 int queue[m*n][2];
55 int front = 0, back = 0;
56 queue[back][0] = sr;
57 queue[back++][1] = sc;
58 int steps = 0;
59 while (front < back) {
60 int size = back - front;
61 steps++;
62 for (int i = 0; i < size; i++) {
63 int r = queue[front][0];
64 int c = queue[front][1];
65 front++;
66 for (int d = 0; d < 4; d++) {
67 int nr = r + directions[d][0];
68 int nc = c + directions[d][1];
69 if (nr >= 0 && nr < m && nc >= 0 && nc < n && !visited[nr][nc] && forest[nr][nc] > 0) {
70 if (nr == tr && nc == tc) return steps;
71 visited[nr][nc] = true;
72 queue[back][0] = nr;
73 queue[back++][1] = nc;
74 }
75 }
76 }
77 }
78 return -1;
79}
80
81int cmpfunc(const void* a, const void* b) {
82 int l = ((int*)a)[0];
83 int r = ((int*)b)[0];
84 return (l - r);
85}This C solution first uses BFS to calculate the shortest path between trees, starting from (0, 0) and moving to trees in increasing order of their height. The binary search is used to sort the trees based on their height first. If any tree is unreachable during a BFS computation, the function returns -1.
Using Bidirectional BFS can help in optimizing the pathfinding step by searching from both the start and the goal concurrently. The aim is to decrease the search space substantially as the two searches meet in the middle. This method is an extension to the regular BFS approach, potentially optimizing the traversal cost considerably under specific configurations.
The potential benefits of bidirectional search in space complexity can be realized when effectively reducing the number of explored nodes, as time complexity ideally shortens from O(b^d) to roughly O(b^(d/2)), where b is the branching factor and d the solution depth.
1from typing import List, Tuple
2
Python realizes the bidirectional search using dual BFS queues for start and target, merging states reached from both ends. It prioritizes advancing the search queue with fewer elements, thereby optimizing space and efficiency. The express serialization of nodes helps manage state transitions seamlessly.