Imprima todas las rutas posibles desde la parte superior izquierda hasta la parte inferior derecha de una array mXn

El problema es imprimir todas las rutas posibles desde la parte superior izquierda hasta la parte inferior derecha de una array mXn con las restricciones de que desde cada celda puede moverse solo hacia la derecha o hacia abajo .

Ejemplos: 

Input : 1 2 3
        4 5 6
Output : 1 4 5 6
         1 2 5 6
         1 2 3 6

Input : 1 2 
        3 4
Output : 1 2 4
         1 3 4

El algoritmo es un algoritmo recursivo simple, desde cada celda primero imprime todas las rutas yendo hacia abajo y luego imprime todas las rutas yendo a la derecha. Haga esto recursivamente para cada celda encontrada.

A continuación se muestra la implementación del algoritmo anterior. 

C++

// C++ program to Print all possible paths from
// top left to bottom right of a mXn matrix
#include<iostream>
 
using namespace std;
 
/* mat:  Pointer to the starting of mXn matrix
   i, j: Current position of the robot (For the first call use 0,0)
   m, n: Dimensions of given the matrix
   pi:   Next index to be filed in path array
   *path[0..pi-1]: The path traversed by robot till now (Array to hold the
                  path need to have space for at least m+n elements) */
void printAllPathsUtil(int *mat, int i, int j, int m, int n, int *path, int pi)
{
    // Reached the bottom of the matrix so we are left with
    // only option to move right
    if (i == m - 1)
    {
        for (int k = j; k < n; k++)
            path[pi + k - j] = *((mat + i*n) + k);
        for (int l = 0; l < pi + n - j; l++)
            cout << path[l] << " ";
        cout << endl;
        return;
    }
 
    // Reached the right corner of the matrix we are left with
    // only the downward movement.
    if (j == n - 1)
    {
        for (int k = i; k < m; k++)
            path[pi + k - i] = *((mat + k*n) + j);
        for (int l = 0; l < pi + m - i; l++)
            cout << path[l] << " ";
        cout << endl;
        return;
    }
 
    // Add the current cell to the path being generated
    path[pi] = *((mat + i*n) + j);
 
    // Print all the paths that are possible after moving down
    printAllPathsUtil(mat, i+1, j, m, n, path, pi + 1);
 
    // Print all the paths that are possible after moving right
    printAllPathsUtil(mat, i, j+1, m, n, path, pi + 1);
 
    // Print all the paths that are possible after moving diagonal
    // printAllPathsUtil(mat, i+1, j+1, m, n, path, pi + 1);
}
 
// The main function that prints all paths from
// top left to bottom right in a matrix 'mat' of size mXn
void printAllPaths(int *mat, int m, int n)
{
    int *path = new int[m+n];
    printAllPathsUtil(mat, 0, 0, m, n, path, 0);
}
 
// Driver program to test above functions
int main()
{
    int mat[2][3] = { {1, 2, 3}, {4, 5, 6} };
    printAllPaths(*mat, 2, 3);
    return 0;
}

Java

import java.util.*;
// Java program to Print all possible paths from
// top left to bottom right of a mXn matrix
public class MatrixTraversal
{
 
 
  
    private static void printMatrix(int mat[][], int m, int n,
                                    int i, int j, List<Integer> list)
    {
          //return if i or j crosses matrix size
        if(i > m || j > n)
            return;
        list.add(mat[i][j]);
        if(i == m && j == n){
            System.out.println(list);
        }
        printMatrix(mat, m, n, i+1, j, list);
        printMatrix(mat, m, n, i, j+1, list);
        list.remove(list.size()-1);
         
         
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int m = 2;
        int n = 3;
        int mat[][] = { { 1, 2, 3 },
                        { 4, 5, 6 } };
          List<Integer> list = new ArrayList<>();
        printMatrix(mat, m-1, n-1, 0, 0, list);
    }
}
//This article is contributed by Harneet

Python3

# Python3 program to Print all possible paths from
# top left to bottom right of a mXn matrix
 
'''
/* mat: Pointer to the starting of mXn matrix
i, j: Current position of the robot
     (For the first call use 0, 0)
m, n: Dimensions of given the matrix
pi: Next index to be filed in path array
*path[0..pi-1]: The path traversed by robot till now
                (Array to hold the path need to have
                 space for at least m+n elements) */
'''
def printAllPathsUtil(mat, i, j, m, n, path, pi):
 
    # Reached the bottom of the matrix
    # so we are left with only option to move right
    if (i == m - 1):
        for k in range(j, n):
            path[pi + k - j] = mat[i][k]
 
        for l in range(pi + n - j):
            print(path[l], end = " ")
        print()
        return
 
    # Reached the right corner of the matrix
    # we are left with only the downward movement.
    if (j == n - 1):
 
        for k in range(i, m):
            path[pi + k - i] = mat[k][j]
 
        for l in range(pi + m - i):
            print(path[l], end = " ")
        print()
        return
 
    # Add the current cell
    # to the path being generated
    path[pi] = mat[i][j]
 
    # Print all the paths
    # that are possible after moving down
    printAllPathsUtil(mat, i + 1, j, m, n, path, pi + 1)
 
    # Print all the paths
    # that are possible after moving right
    printAllPathsUtil(mat, i, j + 1, m, n, path, pi + 1)
 
    # Print all the paths
    # that are possible after moving diagonal
    # printAllPathsUtil(mat, i+1, j+1, m, n, path, pi + 1);
 
# The main function that prints all paths
# from top left to bottom right
# in a matrix 'mat' of size mXn
def printAllPaths(mat, m, n):
 
    path = [0 for i in range(m + n)]
    printAllPathsUtil(mat, 0, 0, m, n, path, 0)
 
# Driver Code
mat = [[1, 2, 3],
       [4, 5, 6]]
 
printAllPaths(mat, 2, 3)
 
# This code is contributed by Mohit Kumar

C#

// C# program to Print all possible paths from
// top left to bottom right of a mXn matrix
using System;
     
public class MatrixTraversal
{
 
 
    /* mat: Pointer to the starting of mXn matrix
i, j: Current position of the robot (For the first call use 0,0)
m, n: Dimensions of given the matrix
pi: Next index to be filed in path array
*path[0..pi-1]: The path traversed by robot till now (Array to hold the
                path need to have space for at least m+n elements) */
    private static void printMatrix(int [,]mat, int m, int n,
                                    int i, int j, int []path, int idx)
    {
        path[idx] = mat[i,j];
         
        // Reached the bottom of the matrix so we are left with
        // only option to move right
        if (i == m - 1)
        {
            for (int k = j + 1; k < n; k++)
            {
                path[idx + k - j] = mat[i,k];
            }
            for (int l = 0; l < idx + n - j; l++)
            {
                Console.Write(path[l] + " ");
            }
            Console.WriteLine();
            return;
        }
         
        // Reached the right corner of the matrix we are left with
        // only the downward movement.
        if (j == n - 1)
        {
            for (int k = i + 1; k < m; k++)
            {
                path[idx + k - i] = mat[k,j];
            }
            for (int l = 0; l < idx + m - i; l++)
            {
                Console.Write(path[l] + " ");
            }
            Console.WriteLine();
            return;
        }
         
        // Print all the paths that are possible after moving down
        printMatrix(mat, m, n, i + 1, j, path, idx + 1);
 
        // Print all the paths that are possible after moving right
        printMatrix(mat, m, n, i, j + 1, path, idx + 1);
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int m = 2;
        int n = 3;
        int [,]mat = { { 1, 2, 3 },
                        { 4, 5, 6 } };
        int maxLengthOfPath = m + n - 1;
        printMatrix(mat, m, n, 0, 0, new int[maxLengthOfPath], 0);
    }
}
 
// This code contributed by Rajput-Ji

Javascript

<script>
// Javascript program to Print all possible paths from
// top left to bottom right of a mXn matrix
 
/* mat:  Pointer to the starting of mXn matrix
   i, j: Current position of the robot (For the first call use 0,0)
   m, n: Dimensions of given the matrix
   pi:   Next index to be filed in path array
   *path[0..pi-1]: The path traversed by robot till now (Array to hold the
                  path need to have space for at least m+n elements) */
function printMatrix(mat,m,n,i,j,path,idx)
{
    path[idx] = mat[i][j];
          
         // Reached the bottom of the matrix so we are left with
        // only option to move right
        if (i == m - 1)
        {
            for (let k = j + 1; k < n; k++)
            {
                path[idx + k - j] = mat[i][k];
            }
            for (let l = 0; l < idx + n - j; l++)
            {
                document.write(path[l] + " ");
            }
            document.write("<br>");
            return;
        }
          
        // Reached the right corner of the matrix we are left with
        // only the downward movement.
        if (j == n - 1)
        {
            for (let k = i + 1; k < m; k++)
            {
                path[idx + k - i] = mat[k][j];
            }
            for (let l = 0; l < idx + m - i; l++)
            {
                document.write(path[l] + " ");
            }
            document.write();
            return;
        }
        // Print all the paths that are possible after moving down
        printMatrix(mat, m, n, i + 1, j, path, idx + 1);
  
         // Print all the paths that are possible after moving right
        printMatrix(mat, m, n, i, j + 1, path, idx + 1);
}
 
// Driver code
let m = 2;
let n = 3;
let mat = [[ 1, 2, 3 ],
[ 4, 5, 6 ]];
let maxLengthOfPath = m + n - 1;
printMatrix(mat, m, n, 0, 0, new Array(maxLengthOfPath), 0);
 
 
// This code is contributed by ab2127
</script>
Producción

1 4 5 6 
1 2 5 6 
1 2 3 6 

Tenga en cuenta que en el código anterior, la última línea de printAllPathsUtil() está comentada. Si descomentamos esta línea, obtenemos todas las rutas desde la parte superior izquierda hasta la parte inferior derecha de una array nXm si también se permiten los movimientos diagonales. Y también si no se permite moverse a algunas de las celdas, entonces el mismo código se puede mejorar pasando la array de restricción a la función anterior y eso se deja como ejercicio.

Este artículo es una contribución de Hariprasad NG . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.

C++

// C++ program to Print all possible paths from 
// top left to bottom right of a mXn matrix
#include <bits/stdc++.h>
using namespace std;
 
vector<vector<int>> allPaths;
 
void findPathsUtil(vector<vector<int>> maze, int m,
                                 int n, int i, int j,
                          vector<int> path, int index)
{
     
    // If we reach the bottom of maze,
    // we can only move right
    if (i == m - 1)
    {
        for(int k = j; k < n; k++)
        {
             
            //path.append(maze[i][k])
            path[index + k - j] = maze[i][k];
        }
         
        // If we hit this block, it means one
        // path is completed. Add it to paths
        // list and print
        cout << "[" << path[0] << ", ";
        for(int z = 1; z < path.size() - 1; z++)
        {
            cout << path[z] << ", ";
        }
        cout << path[path.size() - 1] << "]" << endl;
        allPaths.push_back(path);
        return;
    }
         
    // If we reach to the right most
    // corner, we can only move down
    if (j == n - 1)
    {
        for(int k = i; k < m; k++)
        {
            path[index + k - i] = maze[k][j];
        }
         
        //path.append(maze[j][k])
        // If we hit this block, it means one
        // path is completed. Add it to paths
        // list and print
        cout << "[" << path[0] << ", ";
        for(int z = 1; z < path.size() - 1; z++)
        {
            cout << path[z] << ", ";
        }
        cout << path[path.size() - 1] << "]" << endl;
        allPaths.push_back(path);
        return;
    }
       
    // Add current element to the path list
    //path.append(maze[i][j])
    path[index] = maze[i][j];
       
    // Move down in y direction and call
    // findPathsUtil recursively
    findPathsUtil(maze, m, n, i + 1,
                  j, path, index + 1);
       
    // Move down in y direction and
    // call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j + 1,
                        path, index + 1);
}
     
void findPaths(vector<vector<int>> maze,
                       int m, int n)
{
    vector<int> path(m + n - 1, 0);
    findPathsUtil(maze, m, n, 0, 0, path, 0);
}
 
// Driver Code
int main()
{
    vector<vector<int>> maze{ { 1, 2, 3 },
                              { 4, 5, 6 },
                              { 7, 8, 9 } };
    findPaths(maze, 3, 3);
     
    //print(allPaths)
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program to Print all possible paths from 
// top left to bottom right of a mXn matrix
import java.io.*;
import java.util.*;
class GFG
{
  static ArrayList<ArrayList<Integer>> allPaths =
    new ArrayList<ArrayList<Integer>>();
  static void findPathsUtil(ArrayList<ArrayList<Integer>> maze,
                            int m, int n, int i,int j,
                            ArrayList<Integer> path,int index)
  {
 
    // If we reach the bottom of maze,
    // we can only move right
    if(i == m - 1)
    {
      for(int k = j; k < n; k++)
      {
 
        // path.append(maze[i][k])
        path.set(index + k - j, maze.get(i).get(k));
 
      }
 
      // If we hit this block, it means one
      // path is completed. Add it to paths
      // list and print
      System.out.print("[" + path.get(0) + ", ");
      for(int z = 1; z < path.size() - 1; z++)
      {
        System.out.print(path.get(z) + ", ");
      }
      System.out.println(path.get(path.size() - 1) + "]");
      allPaths.add(path);
      return;
    }
 
    // If we reach to the right most
    // corner, we can only move down
    if(j == n - 1)
    {
      for(int k = i; k < m; k++)
      {
        path.set(index + k - i,maze.get(k).get(j));
      }
 
      // path.append(maze[j][k])
      // If we hit this block, it means one
      // path is completed. Add it to paths
      // list and print
      System.out.print("[" + path.get(0) + ", ");
      for(int z = 1; z < path.size() - 1; z++)
      {
        System.out.print(path.get(z) + ", ");
 
      }
      System.out.println(path.get(path.size() - 1) + "]");
      allPaths.add(path);
      return;
    }
 
    // Add current element to the path list
    //path.append(maze[i][j])
    path.set(index,maze.get(i).get(j));
 
    // Move down in y direction and call
    // findPathsUtil recursively
    findPathsUtil(maze, m, n, i + 1, j, path, index + 1);
 
    // Move down in y direction and
    // call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j + 1, path, index + 1);
 
  }
  static void findPaths(ArrayList<ArrayList<Integer>> maze,
                        int m, int n)
  {
    ArrayList<Integer> path = new ArrayList<Integer>();
    for(int i = 0; i < m + n - 1; i++)
    {
      path.add(0);
    }
    findPathsUtil(maze, m, n, 0, 0, path, 0);
  }
 
  // Driver code
  public static void main (String[] args)
  {
    ArrayList<ArrayList<Integer>> maze =
      new ArrayList<ArrayList<Integer>>();
    maze.add(new ArrayList<Integer>
             (Arrays.asList(1,2,3)));
    maze.add(new ArrayList<Integer>
             (Arrays.asList(4,5,6)));
    maze.add(new ArrayList<Integer>
             (Arrays.asList(7,8,9)));
 
    findPaths(maze, 3, 3);       
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to Print all possible paths from
# top left to bottom right of a mXn matrix
allPaths = []
def findPaths(maze,m,n):
    path = [0 for d in range(m+n-1)]
    findPathsUtil(maze,m,n,0,0,path,0)
     
def findPathsUtil(maze,m,n,i,j,path,index):
    global allPaths
    # if we reach the bottom of maze, we can only move right
    if i==m-1:
        for k in range(j,n):
            #path.append(maze[i][k])
            path[index+k-j] = maze[i][k]
        # if we hit this block, it means one path is completed.
        # Add it to paths list and print
        print(path)
        allPaths.append(path)
        return
    # if we reach to the right most corner, we can only move down
    if j == n-1:
        for k in range(i,m):
            path[index+k-i] = maze[k][j]
          #path.append(maze[j][k])
        # if we hit this block, it means one path is completed.
        # Add it to paths list and print
        print(path)
        allPaths.append(path)
        return
     
    # add current element to the path list
    #path.append(maze[i][j])
    path[index] = maze[i][j]
     
    # move down in y direction and call findPathsUtil recursively
    findPathsUtil(maze, m, n, i+1, j, path, index+1)
     
    # move down in y direction and call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j+1, path, index+1)
 
if __name__ == '__main__':
    maze = [[1,2,3],
            [4,5,6],
            [7,8,9]]
    findPaths(maze,3,3)
    #print(allPaths)

C#

// C# program to Print all possible paths from 
// top left to bottom right of a mXn matrix
using System;
using System.Collections.Generic;
class GFG
{
   
    static List<List<int>> allPaths = new List<List<int>>();
  
    static void findPathsUtil(List<List<int>> maze,
                              int m, int n, int i,
                              int j, List<int> path,
                              int index)
    {
          
        // If we reach the bottom of maze,
        // we can only move right
        if (i == m - 1)
        {
            for(int k = j; k < n; k++)
            {
                  
                // path.append(maze[i][k])
                path[index + k - j] = maze[i][k];
            }
              
            // If we hit this block, it means one
            // path is completed. Add it to paths
            // list and print
            Console.Write( "[" + path[0] + ", ");
            for(int z = 1; z < path.Count - 1; z++)
            {
                Console.Write(path[z] + ", ");
            }
            Console.WriteLine(path[path.Count - 1] + "]");
            allPaths.Add(path);
            return;
        }
              
        // If we reach to the right most
        // corner, we can only move down
        if (j == n - 1)
        {
            for(int k = i; k < m; k++)
            {
                path[index + k - i] = maze[k][j];
            }
              
            // path.append(maze[j][k])
            // If we hit this block, it means one
            // path is completed. Add it to paths
            // list and print
            Console.Write( "[" + path[0] + ", ");
            for(int z = 1; z < path.Count - 1; z++)
            {
                Console.Write(path[z] + ", ");
            }
            Console.WriteLine(path[path.Count - 1] + "]");
            allPaths.Add(path); 
            return;
        }
            
        // Add current element to the path list
        //path.append(maze[i][j])
        path[index] = maze[i][j];
            
        // Move down in y direction and call
        // findPathsUtil recursively
        findPathsUtil(maze, m, n, i + 1,
                      j, path, index + 1);
            
        // Move down in y direction and
        // call findPathsUtil recursively
        findPathsUtil(maze, m, n, i, j + 1,
                            path, index + 1);
    }
          
    static void findPaths(List<List<int>> maze, int m, int n)
    {
        List<int> path = new List<int>();
        for(int i = 0; i < m + n - 1; i++)
        {
            path.Add(0);
        }
        findPathsUtil(maze, m, n, 0, 0, path, 0);
    }
 
  // Driver code
  static void Main()
  {
    List<List<int>> maze = new List<List<int>>();
    maze.Add(new List<int> { 1, 2, 3 });
    maze.Add(new List<int> { 4, 5, 6 });
    maze.Add(new List<int> { 7, 8, 9 });
 
    findPaths(maze, 3, 3);
  }
}
 
// This code is contributed by divyesh072019

Javascript

<script>
// Javascript program to Print all possible paths from 
// top left to bottom right of a mXn matrix
let allPaths = [];
 
function findPathsUtil(maze, m, n, i, j, path, index)
{
     
    // If we reach the bottom of maze,
    // we can only move right
    if (i == m - 1)
    {
        for(let k = j; k < n; k++)
        {
             
            //path.append(maze[i][k])
            path[index + k - j] = maze[i][k];
        }
         
        // If we hit this block, it means one
        // path is completed. Add it to paths
        // list and print
        document.write("[" + path[0] + ", ");
        for(let z = 1; z < path.length - 1; z++)
        {
            document.write(path[z] + ", ");
        }
        document.write(path[path.length - 1] + "]" + "<br>");
        allPaths.push(path);
        return;
    }
         
    // If we reach to the right most
    // corner, we can only move down
    if (j == n - 1)
    {
        for(let k = i; k < m; k++)
        {
            path[index + k - i] = maze[k][j];
        }
         
        // path.append(maze[j][k])
        // If we hit this block, it means one
        // path is completed. Add it to paths
        // list and print
        document.write("[" + path[0] + ", ");
        for(let z = 1; z < path.length - 1; z++)
        {
            document.write(path[z] + ", ");
        }
        document.write(path[path.length - 1] + "]" + "<br>");
        allPaths.push(path);
        return;
    }
       
    // Add current element to the path list
    // path.append(maze[i][j])
    path[index] = maze[i][j];
       
    // Move down in y direction and call
    // findPathsUtil recursively
    findPathsUtil(maze, m, n, i + 1,
                  j, path, index + 1);
       
    // Move down in y direction and
    // call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j + 1,
                        path, index + 1);
}
     
function findPaths(maze, m, n)
{
    let path = new Array(m + n - 1).fill(0);
    findPathsUtil(maze, m, n, 0, 0, path, 0);
}
 
// Driver Code
let maze = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ];
findPaths(maze, 3, 3)
 
// This code is contributed by Saurabh Jaiswal
</script>
Producción

[1, 4, 7, 8, 9]
[1, 4, 5, 8, 9]
[1, 4, 5, 6, 9]
[1, 2, 5, 8, 9]
[1, 2, 5, 6, 9]
[1, 2, 3, 6, 9]

Tenga en cuenta que todo el enfoque anterior requiere tiempo y espacio adicionales para resolver el problema, simplemente podemos usar el algoritmo de seguimiento para resolver el problema de manera optimizada 

C++

#include<bits/stdc++.h>
using namespace std;
 
// function to display the path
void display(vector<int> &ans) {
  for(auto i :ans ) {
    cout<<i <<" ";
  }
  cout<<endl;
}
 
// a function which check whether our step is safe or not
bool issafe(int r,int c,vector<vector<int>>& visited,int n,int m) {
  return (r < n and c <m and visited[r] !=-1 );  // return true if all values satisfied else false
}
 
 
void FindPaths(vector<vector<int>> &grid,int r,int c, int n,int m,vector<int> &ans) {
  // when we hit the last cell we reach to destination then directly push the path
  if(r == n-1 and c == m-1) {
    ans.push_back(grid[r]);
    display(ans);  // function to display the path stored in ans vector
    ans.pop_back(); // pop back because we need to backtrack to explore more path
    return ;
  } 
   
  // we will store the current value in ch and mark the visited place as -1
  int ch = grid[r];
  
  ans.push_back(ch); // push the path in ans array
  grid[r] = -1;  // mark the visited place with -1
   
  // if is it safe to take next downward step then take it
  if(issafe(r+1,c,grid,n,m)) {
    FindPaths(grid,r+1,c,n,m,ans);
  }
   
  // if is it safe to take next rightward step then take it
  if(issafe(r,c+1,grid,n,m)) {
    FindPaths(grid,r,c+1,n,m,ans);
  }
   
  // backtracking step we need to make values original so to we can visit it by some another path
  grid[r] = ch;
   
  // remove the current path element we explore
  ans.pop_back();
  return ;
}
 
int main() {
      int n = 3 ,m =3;
      vector<vector<int> >grid{ {1,2,3},{4,5,6},{7,8,9}};
      vector<int>ans ; // it will store the path which we have covered
       
      FindPaths(grid,0,0,n,m,ans); // here 0,0 are initial position to start with
    return 0;
}

Java

import java.util.*;
 
class Main {
    public static void main(String[] args)
    {
        int n = 3, m = 3;
        int[][] grid
            = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
        ArrayList<Integer> ans
            = new ArrayList<>(); // it will store the path
                                 // which we have covered
 
        FindPaths(grid, 0, 0, n, m,
                  ans); // here 0,0 are initial position to
                        // start with
    }
    // function to display the path
    public static void display(ArrayList<Integer> ans)
    {
        for (int i : ans) {
            System.out.print(i + " ");
        }
        System.out.println();
    }
    // a function which check whether our step is safe or
    // not
    public static boolean
    issafe(int r, int c, int[][] visited, int n, int m)
    {
        return (r < n && c < m && visited[r] != -1);
    }
 
    public static void FindPaths(int[][] grid, int r, int c,
                                 int n, int m,
                                 ArrayList<Integer> ans)
    {
        // when we hit the last cell we reach to destination
        // then directly push the path
        if (r == n - 1 && c == m - 1) {
            ans.add(grid[r]);
            display(ans); // function to display the path
                          // stored in ans vector
            ans.remove(
                ans.size()
                - 1); // remove last element because we need
                      // to backtrack to explore more path
            return;
        }
        // we will store the current value in ch and mark
        // the visited place as -1
        int ch = grid[r];
 
        ans.add(ch); // add the path in ans array
        grid[r] = -1; // mark the visited place with -1
 
        // if is it safe to take next downward step then
        // take it
        if (issafe(r + 1, c, grid, n, m)) {
            FindPaths(grid, r + 1, c, n, m, ans);
        }
        // if is it safe to take next rightward step then
        // take it
        if (issafe(r, c + 1, grid, n, m)) {
            FindPaths(grid, r, c + 1, n, m, ans);
        }
        // backtracking step we need to make values original
        // so to we can visit it by some another path
        grid[r] = ch;
        // remove the current path element we explore
        ans.remove(ans.size() - 1);
        return;
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)

Python3

# code
class Solution:
     
    def __init__(self):
        self.mapping = {}
     
    def printAllPaths(self, M, m, n):
        if not self.mapping.get((m,n)):
            if m == 1 and n == 1:
                return [M[m-1][n-1]]
            else:
                res = []
                if n > 1:
                    a = self.printAllPaths(M, m, n-1)
                    for i in a:
                        if not isinstance(i, list):
                            i = [i]
                        res.append(i+[M[m-1][n-1]])
                if m > 1:
                    b = self.printAllPaths(M, m-1, n)
                    for i in b:
                        if not isinstance(i, list):
                            i = [i]
                        res.append(i+[M[m-1][n-1]])
            self.mapping[(m,n)] = res
        return self.mapping.get((m,n))
 
M = [[1, 2, 3], [4, 5, 6], [7,8,9]]
m, n = len(M), len(M[0])
a = Solution()
res = a.printAllPaths(M, m, n)
for i in res:
    print(i)

C#

// C# program for above problem
using System;
using System.Collections.Generic;
 
class Program {
 
    // function to display the path
    static void display(List<int> ans)
    {
        foreach(var i in ans)
        {
            Console.Write(i + " ");
             
        }
 
        Console.WriteLine();
    }
 
    // a function which check whether our step is safe or not
    static bool issafe(int r, int c, int[][] visited, int n, int m)
    {
        return (r < n && c < m && visited[r] != -1); // return true if all values satisfied else false
    }
 
    static void FindPaths(int[][] grid, int r, int c, int n, int m, List<int> ans)
    {
        // when we hit the last cell we reach to destination then directly push the path
        if (r == n - 1 && c == m - 1) {
            ans.Add(grid[r]);
            display(ans); // function to display the path stored in ans vector
            ans.RemoveAt(ans.Count - 1); // remove last element because we need to backtrack to explore more path
            return;
        }
 
        int ch = grid[r];  // we will store the current value in ch and mark the visited place as -1
 
        ans.Add(ch); // add the path in ans array
 
        grid[r] = -1;  // mark the visited place with -1
 
        // if is it safe to take next downward step then take it
        if (issafe(r + 1, c, grid, n, m)) {
            FindPaths(grid, r + 1, c, n, m, ans);
        }
 
        // if is it safe to take next rightward step then take it
        if (issafe(r, c + 1, grid, n, m)) {
            FindPaths(grid, r, c + 1, n, m, ans);
        }
 
        // backtracking step we need to make values original so to we can visit it by some another path
        grid[r] = ch;
         
        // remove the current path element we explore
        ans.RemoveAt(ans.Count - 1);
    }
 
    static void Main()
    {
        int n = 3, m = 3;
        int[][] grid = new int[n][];
         
        grid[0] = new[] { 1, 2, 3 };
        grid[1] = new[] { 4, 5, 6 };
        grid[2] = new[] { 7, 8, 9 };
         
        List<int> ans = new List<int>(); // it will store the path which we have covered
 
         
        FindPaths(grid, 0, 0, n, m, ans); // here 0,0 are initial position to start with
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)
Producción

1 4 7 8 9 
1 4 5 8 9 
1 4 5 6 9 
1 2 5 8 9 
1 2 5 6 9 
1 2 3 6 9 

Entonces, con este método, puede optimizar su código.

TC- O(2^n*m)    , SC – O(n) 

Otro enfoque (iterativo):

1. En este enfoque, utilizaremos BFS (búsqueda primero en amplitud) para encontrar todas las rutas posibles.

2. Haremos una cola que contenga la siguiente información: 

    a) Vector que almacena el camino hasta una determinada celda.

    b) coordenadas de la celda.

3. Comenzaremos desde la celda superior izquierda y empujaremos el valor de la celda y las coordenadas en la cola.

4. Seguiremos explorando la celda derecha y abajo (si es posible) hasta que la cola no esté vacía 

   y colóquelos en la cola actualizando el vector de celda actual.

5. Si llegamos a la última celda, tenemos una respuesta e imprimiremos el vector de respuesta.

C++

// c++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// this structure stores information
// about a particular cell i.e
// path upto that cell and cell's
// coordinates
struct info {
    vector<int> path;
    int i;
    int j;
};
 
void printAllPaths(vector<vector<int> >& maze)
{
    int n = maze.size();
    int m = maze[0].size();
 
    queue<info> q;
    // pushing top-left cell into the queue
    q.push({ { maze[0][0] }, 0, 0 });
 
    while (!q.empty()) {
        info p = q.front();
        q.pop();
 
        // if we reached the bottom-right cell
        // i.e the destination then print the path
        if (p.i == n - 1 && p.j == m - 1) {
            for (auto x : p.path)
                cout << x << " ";
 
            cout << "\n";
        }
 
        // if we are in the last row
        // then only right movement is possible
        else if (p.i == n - 1) {
            vector<int> temp = p.path;
            // updating the current path
            temp.push_back(maze[p.i][p.j + 1]);
 
            q.push({ temp, p.i, p.j + 1 });
        }
 
        // if we are in the last column
        // then only down movement is possible
        else if (p.j == m - 1) {
            vector<int> temp = p.path;
            // updating the current path
            temp.push_back(maze[p.i + 1][p.j]);
 
            q.push({ temp, p.i + 1, p.j });
        }
 
        // else both right and down movement
        // are possible
        else { // right movement
            vector<int> temp = p.path;
            // updating the current path
            temp.push_back(maze[p.i][p.j + 1]);
 
            q.push({ temp, p.i, p.j + 1 });
 
            // down movement
            temp.pop_back();
            // updating the current path
            temp.push_back(maze[p.i + 1][p.j]);
 
            q.push({ temp, p.i + 1, p.j });
        }
    }
}
 
// Driver Code
int main()
{
    vector<vector<int> > maze{ { 1, 2, 3 },
                               { 4, 5, 6 },
                               { 7, 8, 9 } };
 
    printAllPaths(maze);
 
    return 0;
}

Python3

# Python implementation for the above approach
from collections import deque
 
# this structure stores information
# about a particular cell i.e
# path upto that cell and cell's
# coordinates
 
 
class info:
    def __init__(self, path, i, j):
        self.path = path
        self.i = i
        self.j = j
 
 
def printAllPaths(maze):
    n = len(maze)
    m = len(maze[0])
 
    q = deque()
    # pushing top-left cell into the queue
    q.append(info([maze[0][0]], 0, 0))
 
    while len(q) > 0:
        p = q.popleft()
 
        # if we reached the bottom-right cell
        # i.e the destination then print the path
        if p.i == n - 1 and p.j == m - 1:
            for x in p.path:
                print(x, end=" ")
            print()
 
        # if we are in the last row
        # then only right movement is possible
        elif p.i == n-1:
            temp = p.path[:]
            # updating the current path
            temp.append(maze[p.i][p.j+1])
            q.append(info(temp, p.i, p.j+1))
 
        # if we are in the last column
        # then only down movement is possible
        elif p.j == m-1:
            temp = p.path[:]
            # updating the current path
            temp.append(maze[p.i+1][p.j])
            q.append(info(temp, p.i+1, p.j))
 
        # else both right and down movement
        # are possible
        else:  # right movement
            temp = p.path[:]
            # updating the current path
            temp.append(maze[p.i][p.j + 1])
            q.append(info(temp, p.i, p.j + 1))
 
            # down movement
            temp = temp[:-1]
 
            # updating the current path
            temp.append(maze[p.i + 1][p.j])
            q.append(info(temp, p.i + 1, p.j))
 
 
# Driver Code
maze = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
printAllPaths(maze)
 
# This code is contributed by Tapesh(tapeshdua420)
Producción

1 2 3 6 9 
1 2 5 6 9 
1 2 5 8 9 
1 4 5 6 9 
1 4 5 8 9 
1 4 7 8 9 

  

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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