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מצא את המסלול הבטוח הקצר ביותר בנתיב עם מוקשים

נסה את זה ב-GfG Practice input_matrix' title=

נתון מטריצה ​​מלבנית mat[][] כאשר חלק מהתאים מכילים מוקשים (מסומנים ב-0) והשאר בטוחים (מסומנים ב-1) מצאו את אורך המסלול הבטוח הקצר ביותר מכל תא בעמודה הראשונה לכל תא בעמודה האחרונה.

  • מוקשים אינם בטוחים וארבעת התאים הסמוכים שלהם (למעלה למטה משמאל ימין) אינם בטוחים אף הם.
  • רק מעברים אופקיים ואנכיים לתאים בטוחים סמוכים מותרים.
  • אם אי אפשר להגיע לעמוד האחרון בבטחה החזר -1.

דוגמאות:  



מחולל ערכים אקראי ב-java

קֶלֶט:
עם[][] = [ [1 0 1 1 1]
[1 1 1 1 1]
[1 1 1 1 1]
[1 1 1 0 1]
[1 1 1 1 0] ]
תְפוּקָה: 6
הֶסבֵּר:

' title=

אנחנו יכולים לראות את האורך הקצר ביותר
מסלול בטוח הוא 6.

קֶלֶט:
עם[][] = [ [1 1 1 1 1]
[1 1 0 1 1]
[1 1 1 1 1] ]
תְפוּקָה: -1
הֶסבֵּר: אין דרך אפשרית מ
עמודה ראשונה לעמודה אחרונה.

תוכן עניינים



היא מערכת יחסים

[גישה] שימוש ב-backtracking

הרעיון הוא להשתמש חזרה לאחור . תחילה אנו מסמנים את כל התאים הסמוכים של המוקשים כלא בטוחים. ואז עבור כל תא בטוח של העמודה הראשונה של המטריצה ​​אנו מתקדמים לכל הכיוונים המותרים ובודקים באופן רקורסיבי אם הם מובילים ליעד או לא. אם נמצא יעד אנו מעדכנים את הערך של הנתיב הקצר ביותר, אחרת אם אף אחד מהפתרונות שלמעלה לא עובד, נחזיר false מהפונקציה שלנו.

C++
#include    #include  #include  #include    using namespace std; // Function to mark unsafe cells (landmines and their adjacent cells) void markUnsafeCells(vector<vector<int>> &mat) {  int r = mat.size();  int c = mat[0].size();    // Directions for adjacent cells: up down left right  int row[] = {-1 1 0 0};  int col[] = {0 0 -1 1};    vector<vector<int>> temp = mat;    // Mark adjacent cells of landmines (0) as unsafe (0)  for (int i = 0; i < r; i++) {  for (int j = 0; j < c; j++) {  if (temp[i][j] == 0) {  for (int k = 0; k < 4; k++) {  int ni = i + row[k];  int nj = j + col[k];  if (ni >= 0 && ni < r && nj >= 0 && nj < c) {  mat[ni][nj] = 0;  }  }  }  }  } } // DFS to find shortest path from (i j) to any cell in last column int dfs(vector<vector<int>> &mat vector<vector<bool>> &visited int i int j int c) {  int r = mat.size();    if (i < 0 || i >= r || j < 0 || j >= c || mat[i][j] == 0 || visited[i][j]) {  return INT_MAX;  }    if (j == c - 1) {  return 1;  }    visited[i][j] = true;    // Four possible moves: up down left right  int row[] = {-1 1 0 0};  int col[] = {0 0 -1 1};    int minPath = INT_MAX;    // Try all four directions  for (int k = 0; k < 4; k++) {  int ni = i + row[k];  int nj = j + col[k];    int pathLength = dfs(mat visited ni nj c);  if (pathLength != INT_MAX) {  minPath = min(minPath 1 + pathLength);  }  }    // Backtrack - unmark current cell  visited[i][j] = false;    return minPath; } int findShortestPath(vector<vector<int>> &mat) {  int r = mat.size();  int c = mat[0].size();    // Mark all adjacent cells of landmines as unsafe  markUnsafeCells(mat);    // Initialize visited array  vector<vector<bool>> visited(r vector<bool>(c false));    int minPath = INT_MAX;    // Try starting from each safe cell in the first column  for (int i = 0; i < r; i++) {  if (mat[i][0] == 1) {  int pathLength = dfs(mat visited i 0 c);  if (pathLength != INT_MAX) {  minPath = min(minPath pathLength);  }  }  }    return minPath == INT_MAX ? -1 : minPath; } int main() {  vector<vector<int>> mat = {  {1 0 1 1 1}  {1 1 1 1 1}  {1 1 1 1 1}  {1 1 1 0 1}  {1 1 1 1 0}  };    int result = findShortestPath(mat);  cout << result << endl;    return 0; } 
Java
import java.util.Arrays; class Solution {  // Function to mark unsafe cells (landmines and their adjacent cells)  private void markUnsafeCells(int[][] mat) {  int r = mat.length;  int c = mat[0].length;      int[] row = {-1 1 0 0};  int[] col = {0 0 -1 1};    // Create a copy to avoid modifying original safe cells prematurely  int[][] temp = new int[r][c];  for (int i = 0; i < r; i++) {  for (int j = 0; j < c; j++) {  temp[i][j] = mat[i][j];  }  }    // Mark adjacent cells of landmines (0) as unsafe (0)  for (int i = 0; i < r; i++) {  for (int j = 0; j < c; j++) {  if (temp[i][j] == 0) {  for (int k = 0; k < 4; k++) {  int ni = i + row[k];  int nj = j + col[k];  if (ni >= 0 && ni < r && nj >= 0 && nj < c) {  mat[ni][nj] = 0;  }  }  }  }  }  }    // DFS to find shortest path from (i j) to any cell in last column  private int dfs(int[][] mat boolean[][] visited int i int j int c) {  int r = mat.length;    // If out of bounds blocked or visited  if (i < 0 || i >= r || j < 0 || j >= c || mat[i][j] == 0 || visited[i][j]) {  return Integer.MAX_VALUE;  }  if (j == c - 1) {  return 1;  }  visited[i][j] = true;    int[] row = {-1 1 0 0};  int[] col = {0 0 -1 1};    int minPath = Integer.MAX_VALUE;    // Try all four directions  for (int k = 0; k < 4; k++) {  int ni = i + row[k];  int nj = j + col[k];    int pathLength = dfs(mat visited ni nj c);  if (pathLength != Integer.MAX_VALUE) {  minPath = Math.min(minPath 1 + pathLength);  }  }    // Backtrack - unmark current cell  visited[i][j] = false;    return minPath;  }    public int findShortestPath(int[][] mat) {  int r = mat.length;  int c = mat[0].length;    // Mark all adjacent cells of landmines as unsafe  markUnsafeCells(mat);    boolean[][] visited = new boolean[r][c];    int minPath = Integer.MAX_VALUE;    // Try starting from each safe cell in the first column  for (int i = 0; i < r; i++) {  if (mat[i][0] == 1) {  int pathLength = dfs(mat visited i 0 c);  if (pathLength != Integer.MAX_VALUE) {  minPath = Math.min(minPath pathLength);  }  }  }    return minPath == Integer.MAX_VALUE ? -1 : minPath;  }  public static void main(String[] args) {  int[][] mat = {  {1 0 1 1 1}  {1 1 1 1 1}  {1 1 1 1 1}  {1 1 1 0 1}  {1 1 1 1 0}  };    Solution solution = new Solution();  int result = solution.findShortestPath(mat);  System.out.println(result);  } } 
Python
# Function to mark unsafe cells (landmines and their adjacent cells) def mark_unsafe_cells(mat): r = len(mat) c = len(mat[0]) # Directions for adjacent cells: up down left right row = [-1 1 0 0] col = [0 0 -1 1] # Create a copy to avoid modifying original safe cells prematurely temp = [row[:] for row in mat] # Mark adjacent cells of landmines (0) as unsafe (0) for i in range(r): for j in range(c): if temp[i][j] == 0: for k in range(4): ni = i + row[k] nj = j + col[k] if 0 <= ni < r and 0 <= nj < c: mat[ni][nj] = 0 # DFS to find shortest path from (i j) to any cell in last column def dfs(mat visited i j c): r = len(mat) # If out of bounds blocked or visited if i < 0 or i >= r or j < 0 or j >= c or mat[i][j] == 0 or visited[i][j]: return float('inf') if j == c - 1: return 1 visited[i][j] = True # Four possible moves: up down left right row = [-1 1 0 0] col = [0 0 -1 1] min_path = float('inf') # Try all four directions for k in range(4): ni = i + row[k] nj = j + col[k] path_length = dfs(mat visited ni nj c) if path_length != float('inf'): min_path = min(min_path 1 + path_length) # Backtrack - unmark current cell visited[i][j] = False return min_path def findShortestPath(mat): r = len(mat) c = len(mat[0]) # Mark all adjacent cells of landmines as unsafe mark_unsafe_cells(mat) visited = [[False] * c for _ in range(r)] min_path = float('inf') # Try starting from each safe cell in the first column for i in range(r): if mat[i][0] == 1: path_length = dfs(mat visited i 0 c) if path_length != float('inf'): min_path = min(min_path path_length) return -1 if min_path == float('inf') else min_path def main(): mat = [ [1 0 1 1 1] [1 1 1 1 1] [1 1 1 1 1] [1 1 1 0 1] [1 1 1 1 0] ] result = findShortestPath(mat) print(result) if __name__ == '__main__': main() 
C#
using System; class GFG {  // Function to mark unsafe cells (landmines and their adjacent cells)  private void MarkUnsafeCells(int[][] mat) {  int r = mat.Length;  int c = mat[0].Length;    // Directions for adjacent cells: up down left right  int[] row = { -1 1 0 0 };  int[] col = { 0 0 -1 1 };    // Create a copy to avoid modifying original safe cells prematurely  int[][] temp = new int[r][];  for (int i = 0; i < r; i++) {  temp[i] = new int[c];  Array.Copy(mat[i] temp[i] c);  }    // Mark adjacent cells of landmines (0) as unsafe (0)  for (int i = 0; i < r; i++) {  for (int j = 0; j < c; j++) {  if (temp[i][j] == 0) {  for (int k = 0; k < 4; k++) {  int ni = i + row[k];  int nj = j + col[k];  if (ni >= 0 && ni < r && nj >= 0 && nj < c) {  mat[ni][nj] = 0;  }  }  }  }  }  }    // DFS to find shortest path from (i j) to any cell in last column  private int Dfs(int[][] mat bool[][] visited int i int j int c) {  int r = mat.Length;    // If out of bounds blocked or visited  if (i < 0 || i >= r || j < 0 || j >= c || mat[i][j] == 0 || visited[i][j]) {  return int.MaxValue;  }    if (j == c - 1) {  return 1;  }    visited[i][j] = true;  int[] row = { -1 1 0 0 };  int[] col = { 0 0 -1 1 };    int minPath = int.MaxValue;    // Try all four directions  for (int k = 0; k < 4; k++) {  int ni = i + row[k];  int nj = j + col[k];    int pathLength = Dfs(mat visited ni nj c);  if (pathLength != int.MaxValue) {  minPath = Math.Min(minPath 1 + pathLength);  }  }    // Backtrack - unmark current cell  visited[i][j] = false;    return minPath;  }    public int FindShortestPath(int[][] mat) {  int r = mat.Length;  int c = mat[0].Length;    // Mark all adjacent cells of landmines as unsafe  MarkUnsafeCells(mat);    bool[][] visited = new bool[r][];  for (int i = 0; i < r; i++) {  visited[i] = new bool[c];  }    int minPath = int.MaxValue;    // Try starting from each safe cell in the first column  for (int i = 0; i < r; i++) {  if (mat[i][0] == 1) {  int pathLength = Dfs(mat visited i 0 c);  if (pathLength != int.MaxValue) {  minPath = Math.Min(minPath pathLength);  }  }  }    return minPath == int.MaxValue ? -1 : minPath;  }  static void Main(string[] args) {  int[][] mat = new int[][] {  new int[] { 1 0 1 1 1 }  new int[] { 1 1 1 1 1 }  new int[] { 1 1 1 1 1 }  new int[] { 1 1 1 0 1 }  new int[] { 1 1 1 1 0 }  };    GFG solution = new GFG();  int result = solution.FindShortestPath(mat);  Console.WriteLine(result);  } } 
JavaScript
function markUnsafeCells(mat) {  const r = mat.length;  const c = mat[0].length;    // Directions for adjacent cells: up down left right  const row = [-1 1 0 0];  const col = [0 0 -1 1];    // Create a copy to avoid modifying original safe cells prematurely  const temp = mat.map(row => [...row]);    // Mark adjacent cells of landmines (0) as unsafe (0)  for (let i = 0; i < r; i++) {  for (let j = 0; j < c; j++) {  if (temp[i][j] === 0) {  for (let k = 0; k < 4; k++) {  const ni = i + row[k];  const nj = j + col[k];  if (ni >= 0 && ni < r && nj >= 0 && nj < c) {  mat[ni][nj] = 0;  }  }  }  }  } } function dfs(mat visited i j c) {  const r = mat.length;    // If out of bounds blocked or visited  if (i < 0 || i >= r || j < 0 || j >= c || mat[i][j] === 0 || visited[i][j]) {  return Infinity;  }    // If reached the last column  if (j === c - 1) {  return 1;  }    visited[i][j] = true;    const row = [-1 1 0 0];  const col = [0 0 -1 1];    let minPath = Infinity;    // Try all four directions  for (let k = 0; k < 4; k++) {  const ni = i + row[k];  const nj = j + col[k];    const pathLength = dfs(mat visited ni nj c);  if (pathLength !== Infinity) {  minPath = Math.min(minPath 1 + pathLength);  }  }    // Backtrack - unmark current cell  visited[i][j] = false;    return minPath; } function findShortestPath(mat) {  const r = mat.length;  const c = mat[0].length;    // Mark all adjacent cells of landmines as unsafe  markUnsafeCells(mat);    const visited = Array(r).fill().map(() => Array(c).fill(false));    let minPath = Infinity;    // Try starting from each safe cell in the first column  for (let i = 0; i < r; i++) {  if (mat[i][0] === 1) {  const pathLength = dfs(mat visited i 0 c);  if (pathLength !== Infinity) {  minPath = Math.min(minPath pathLength);  }  }  }    return minPath === Infinity ? -1 : minPath; } const mat = [  [1 0 1 1 1]  [1 1 1 1 1]  [1 1 1 1 1]  [1 1 1 0 1]  [1 1 1 1 0] ]; const result = findShortestPath(mat); console.log(result); 

תְפוּקָה
6 

מורכבות זמן: O(4^(r * c)) כאשר r הוא מספר השורות ו-c הם מספר העמודות במטריצה ​​הנתונה.
מרחב עזר: O(r * c) מכיוון שאנו משתמשים בשטח נוסף כמו visted[r][c].

קרא מקובץ csv ב-java

[גישה אופטימלית] שימוש בחיפוש רוחב ראשון

ניתן לפתור את זה בזמן פולינום באמצעות Breadth-First Search. תור את כל התאים הבטוחים מהעמודה האחרונה אל התור עם מרחק = 1. ככל שה-BFS מתקדם, הנתיב הקצר ביותר לכל תא מהעמודה האחרונה מחושב. לבסוף בין כל התאים הנגישים בעמודה הראשונה פלט את המרחק המינימלי.



C++
#include    #include  #include    #include  #include  #include     using namespace std;  int rowNum[4] = {-1 0 0 1};   int colNum[4] = {0 -1 1 0};     int findShortestPath(vector<vector<int>> &mat)  {  int n = mat.size();   int m = mat[0].size();     queue<array<int3>> q; // Queue to perform BFS    int d[n][m];       for(int i = 0; i < n; i++)  for(int j = 0; j < m; j++)  d[i][j] = 1e9;    // Lambda function to check if cell is valid  auto isValid = [&](int i int j) {  if(i < 0 || i >= n || j < 0 || j >= m) return false;  return true;  };    // Lambda function to check if cell and its adjacent cells are safe  auto check = [&](int i int j) {  if(!isValid(i j)) return false;  for(int k = 0; k < 4; k++) {  if(isValid(i + rowNum[k] j + colNum[k]) && !mat[i + rowNum[k]][j + colNum[k]]) return false;  }  return true;  };    // Pushing cells from the rightmost column into the queue  for(int i = 0; i < n; i++) {  if(check(i m - 1)) {  q.push({i m - 1 1});  }  }    // BFS traversal  while(!q.empty()) {  auto z = q.front();  int x = z[0] y = z[1] dis = z[2];  q.pop();  if(d[x][y] > dis) {  d[x][y] = dis;  for(int k = 0; k < 4; k++) {  if(check(x + rowNum[k] y + colNum[k])) {  q.push({x + rowNum[k] y + colNum[k] dis + 1});  }  }  }  }    // Finding the minimum distance in the first column  int ans = 1e9;  for(int i = 0; i < n; i++)  ans = min(ans d[i][0]);    // If no safe path found return -1  if(ans >= 1e9) ans = -1;  return ans;  } int main() {  vector<vector<int>> mat = {  {1 0 1 1 1}  {1 1 1 1 1}  {1 1 1 1 1}  {1 1 1 0 1}  {1 1 1 1 0}  };    int result = findShortestPath(mat);  cout << result << endl;    return 0; } 
Java
import java.util.*; public class Solution {  static int[] rowNum = {-1 0 0 1};  static int[] colNum = {0 -1 1 0};  public static int findShortestPath(int[][] mat) {  int n = mat.length;  int m = mat[0].length;  Queue<int[]> q = new LinkedList<>();  int[][] d = new int[n][m];  // Initializing distance array with large values  for (int i = 0; i < n; i++) {  Arrays.fill(d[i] (int) 1e9);  }  // Lambda-like helper function: check if cell is valid  java.util.function.BiFunction<Integer Integer Boolean> isValid = (i j) -> {  return !(i < 0 || i >= n || j < 0 || j >= m);  };  // Helper function: check if cell and adjacent cells are safe  java.util.function.BiFunction<Integer Integer Boolean> check = (i j) -> {  if (!isValid.apply(i j)) return false;  for (int k = 0; k < 4; k++) {  int ni = i + rowNum[k];  int nj = j + colNum[k];  if (isValid.apply(ni nj) && mat[ni][nj] == 0) return false;  }  return true;  };  // Pushing cells from the rightmost column into the queue  for (int i = 0; i < n; i++) {  if (check.apply(i m - 1)) {  q.add(new int[]{i m - 1 1});  }  }  // BFS traversal  while (!q.isEmpty()) {  int[] z = q.poll();  int x = z[0] y = z[1] dis = z[2];  if (d[x][y] > dis) {  d[x][y] = dis;  for (int k = 0; k < 4; k++) {  int ni = x + rowNum[k];  int nj = y + colNum[k];  if (check.apply(ni nj)) {  q.add(new int[]{ni nj dis + 1});  }  }  }  }  // Finding the minimum distance in the first column  int ans = (int) 1e9;  for (int i = 0; i < n; i++) {  ans = Math.min(ans d[i][0]);  }  // If no safe path found return -1  if (ans >= 1e9) ans = -1;  return ans;  }  public static void main(String[] args) {  int[][] mat = {  {1 0 1 1 1}  {1 1 1 1 1}  {1 1 1 1 1}  {1 1 1 0 1}  {1 1 1 1 0}  };  int result = findShortestPath(mat);  System.out.println(result);  } } 
Python
from collections import deque rowNum = [-1 0 0 1] colNum = [0 -1 1 0] def findShortestPath(mat): n = len(mat) m = len(mat[0]) q = deque() d = [[10**9 for _ in range(m)] for _ in range(n)] # Check if cell is valid def isValid(i j): return not (i < 0 or i >= n or j < 0 or j >= m) # Check if cell and its adjacent cells are safe def check(i j): if not isValid(i j): return False for k in range(4): ni nj = i + rowNum[k] j + colNum[k] if isValid(ni nj) and mat[ni][nj] == 0: return False return True # Pushing cells from the rightmost column into the queue for i in range(n): if check(i m - 1): q.append((i m - 1 1)) # BFS traversal while q: x y dis = q.popleft() if d[x][y] > dis: d[x][y] = dis for k in range(4): ni nj = x + rowNum[k] y + colNum[k] if check(ni nj): q.append((ni nj dis + 1)) # Finding the minimum distance in the first column ans = min(d[i][0] for i in range(n)) # If no safe path found return -1 if ans >= 10**9: ans = -1 return ans if __name__ == '__main__': mat = [ [1 0 1 1 1] [1 1 1 1 1] [1 1 1 1 1] [1 1 1 0 1] [1 1 1 1 0] ] result = findShortestPath(mat) print(result) 
C#
using System; using System.Collections.Generic; class Solution {  static int[] rowNum = { -1 0 0 1 };  static int[] colNum = { 0 -1 1 0 };  // Check if cell is valid  static bool IsValid(int i int j int n int m)  {  return !(i < 0 || i >= n || j < 0 || j >= m);  }  // Check if cell and its adjacent cells are safe  static bool Check(int i int j int[][] mat int n int m)  {  if (!IsValid(i j n m)) return false;  for (int k = 0; k < 4; k++)  {  int ni = i + rowNum[k];  int nj = j + colNum[k];  if (IsValid(ni nj n m) && mat[ni][nj] == 0) return false;  }  return true;  }  public static int FindShortestPath(int[][] mat)  {  int n = mat.Length;  int m = mat[0].Length;  Queue<(int int int)> q = new Queue<(int int int)>();  int[] d = new int[n m];  // Initialize distance array with large value  for (int i = 0; i < n; i++)  for (int j = 0; j < m; j++)  d[i j] = int.MaxValue / 2;  // Push safe cells from rightmost column  for (int i = 0; i < n; i++)  {  if (Check(i m - 1 mat n m))  {  q.Enqueue((i m - 1 1));  }  }  // BFS traversal  while (q.Count > 0)  {  var (x y dis) = q.Dequeue();  if (d[x y] > dis)  {  d[x y] = dis;  for (int k = 0; k < 4; k++)  {  int ni = x + rowNum[k];  int nj = y + colNum[k];  if (Check(ni nj mat n m))  {  q.Enqueue((ni nj dis + 1));  }  }  }  }  // Find minimum distance in the first column  int ans = int.MaxValue / 2;  for (int i = 0; i < n; i++)  ans = Math.Min(ans d[i 0]);  return ans >= int.MaxValue / 2 ? -1 : ans;  }  static void Main()  {  int[][] mat = new int[][]  {  new int[] {1 0 1 1 1}  new int[] {1 1 1 1 1}  new int[] {1 1 1 1 1}  new int[] {1 1 1 0 1}  new int[] {1 1 1 1 0}  };  int result = FindShortestPath(mat);  Console.WriteLine(result);  } } 
JavaScript
function findShortestPath(mat) {  const n = mat.length;  const m = mat[0].length;  const rowNum = [-1 0 0 1];  const colNum = [0 -1 1 0];  // Distance matrix initialized to large value  const d = Array.from({ length: n } () => Array(m).fill(Number.MAX_SAFE_INTEGER));  // Check if cell is valid  function isValid(i j) {  return !(i < 0 || i >= n || j < 0 || j >= m);  }  // Check if cell and its adjacent cells are safe  function check(i j) {  if (!isValid(i j)) return false;  for (let k = 0; k < 4; k++) {  let ni = i + rowNum[k];  let nj = j + colNum[k];  if (isValid(ni nj) && mat[ni][nj] === 0) return false;  }  return true;  }  // Queue for BFS  let q = [];  // Push safe cells from rightmost column  for (let i = 0; i < n; i++) {  if (check(i m - 1)) {  q.push([i m - 1 1]);  }  }  // BFS traversal  while (q.length > 0) {  let [x y dis] = q.shift();  if (d[x][y] > dis) {  d[x][y] = dis;  for (let k = 0; k < 4; k++) {  let ni = x + rowNum[k];  let nj = y + colNum[k];  if (check(ni nj)) {  q.push([ni nj dis + 1]);  }  }  }  }  // Find minimum distance in first column  let ans = Number.MAX_SAFE_INTEGER;  for (let i = 0; i < n; i++) {  ans = Math.min(ans d[i][0]);  }  return ans >= Number.MAX_SAFE_INTEGER ? -1 : ans; } const mat = [  [1 0 1 1 1]  [1 1 1 1 1]  [1 1 1 1 1]  [1 1 1 0 1]  [1 1 1 1 0] ]; const result = findShortestPath(mat); console.log(result); 

תְפוּקָה
6 

מורכבות זמן: O(r * c) כאשר r ו-c הם מספר השורות והעמודות במטריצה ​​הנתונה בהתאמה.
מרחב עזר: O(r * c)