Esta es la variación de Rat en Maze
A Maze se da como N*N array binaria de bloques donde el bloque de origen es el bloque superior izquierdo, es decir, maze[0][0] y el bloque de destino es el bloque inferior derecho, es decir, maze[N -1][N-1]. Una rata parte de la fuente y tiene que llegar al destino. La rata solo puede moverse en dos direcciones: hacia adelante y hacia abajo.
En la array del laberinto, 0 significa que el bloque es un callejón sin salida y un número distinto de cero significa que el bloque se puede utilizar en la ruta desde el origen hasta el destino. El valor distinto de cero de mat[i][j] indica el número máximo de saltos que la rata puede hacer desde la celda mat[i][j].
En esta variación, Rat puede saltar varios pasos a la vez en lugar de 1.
Ejemplos:
Input : { {2, 1, 0, 0},
{3, 0, 0, 1},
{0, 1, 0, 1},
{0, 0, 0, 1}
}
Output : { {1, 0, 0, 0},
{1, 0, 0, 1},
{0, 0, 0, 1},
{0, 0, 0, 1}
}
Explanation
Rat started with M[0][0] and can jump upto 2 steps right/down.
Let's try in horizontal direction -
M[0][1] won't lead to solution and M[0][2] is 0 which is dead end.
So, backtrack and try in down direction.
Rat jump down to M[1][0] which eventually leads to solution.
Input : {
{2, 1, 0, 0},
{2, 0, 0, 1},
{0, 1, 0, 1},
{0, 0, 0, 1}
}
Output : Solution doesn't exist
Algoritmo ingenuo
El algoritmo ingenuo es generar todas las rutas desde el origen hasta el destino y verificar una por una si la ruta generada cumple con las restricciones.
while there are untried paths
{
generate the next path
if this path has all blocks as non-zero
{
print this path;
}
}
Algoritmo de retroceso
If destination is reached
print the solution matrix
Else
a) Mark current cell in solution matrix as 1.
b) Move forward/jump (for each valid steps) in horizontal direction
and recursively check if this move leads to a solution.
c) If the move chosen in the above step doesn't lead to a solution
then move down and check if this move leads to a solution.
d) If none of the above solutions work then unmark this cell as 0
(BACKTRACK) and return false.
Implementación de la solución Backtracking
C++
/* C/C++ program to solve Rat in a Maze problem
using backtracking */
#include <stdio.h>
// Maze size
#define N 4
bool solveMazeUtil(int maze[N][N], int x, int y,
int sol[N][N]);
/* A utility function to print solution matrix
sol[N][N] */
void printSolution(int sol[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf(" %d ", sol[i][j]);
printf("\n");
}
}
/* A utility function to check if x, y is valid
index for N*N maze */
bool isSafe(int maze[N][N], int x, int y)
{
// if (x, y outside maze) return false
if (x >= 0 && x < N && y >= 0 &&
y < N && maze[x][y] != 0)
return true;
return false;
}
/* This function solves the Maze problem using
Backtracking. It mainly uses solveMazeUtil() to
solve the problem. It returns false if no path
is possible, otherwise return true and prints
the path in the form of 1s. Please note that
there may be more than one solutions,
this function prints one of the feasible solutions.*/
bool solveMaze(int maze[N][N])
{
int sol[N][N] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveMazeUtil(maze, 0, 0, sol) == false) {
printf("Solution doesn't exist");
return false;
}
printSolution(sol);
return true;
}
/* A recursive utility function to solve Maze problem */
bool solveMazeUtil(int maze[N][N], int x, int y,
int sol[N][N])
{
// if (x, y is goal) return true
if (x == N - 1 && y == N - 1) {
sol[x][y] = 1;
return true;
}
// Check if maze[x][y] is valid
if (isSafe(maze, x, y) == true) {
// mark x, y as part of solution path
sol[x][y] = 1;
/* Move forward in x direction */
for (int i = 1; i <= maze[x][y] && i < N; i++) {
/* Move forward in x direction */
if (solveMazeUtil(maze, x + i, y, sol) == true)
return true;
/* If moving in x direction doesn't give
solution then Move down in y direction */
if (solveMazeUtil(maze, x, y + i, sol) == true)
return true;
}
/* If none of the above movements work then
BACKTRACK: unmark x, y as part of solution
path */
sol[x][y] = 0;
return false;
}
return false;
}
// driver program to test above function
int main()
{
int maze[N][N] = { { 2, 1, 0, 0 },
{ 3, 0, 0, 1 },
{ 0, 1, 0, 1 },
{ 0, 0, 0, 1 } };
solveMaze(maze);
return 0;
}
Java
// Java program to solve Rat in a Maze problem
// using backtracking
class GFG
{
// Maze size
static int N = 4;
/* A utility function to print solution matrix
sol[N][N] */
static void printSolution(int sol[][])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
System.out.printf(" %d ", sol[i][j]);
}
System.out.printf("\n");
}
}
/* A utility function to check if x, y is valid
index for N*N maze */
static boolean isSafe(int maze[][], int x, int y)
{
// if (x, y outside maze) return false
if (x >= 0 && x < N && y >= 0 &&
y < N && maze[x][y] != 0)
{
return true;
}
return false;
}
/* This function solves the Maze problem using
Backtracking. It mainly uses solveMazeUtil() to
solve the problem. It returns false if no path
is possible, otherwise return true and prints
the path in the form of 1s. Please note that
there may be more than one solutions,
this function prints one of the feasible solutions.*/
static boolean solveMaze(int maze[][])
{
int sol[][] = {{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0}};
if (solveMazeUtil(maze, 0, 0, sol) == false)
{
System.out.printf("Solution doesn't exist");
return false;
}
printSolution(sol);
return true;
}
/* A recursive utility function to solve Maze problem */
static boolean solveMazeUtil(int maze[][], int x,
int y, int sol[][])
{
// if (x, y is goal) return true
if (x == N - 1 && y == N - 1)
{
sol[x][y] = 1;
return true;
}
// Check if maze[x][y] is valid
if (isSafe(maze, x, y) == true)
{
// mark x, y as part of solution path
sol[x][y] = 1;
/* Move forward in x direction */
for (int i = 1; i <= maze[x][y] && i < N; i++)
{
/* Move forward in x direction */
if (solveMazeUtil(maze, x + i, y, sol) == true)
{
return true;
}
/* If moving in x direction doesn't give
solution then Move down in y direction */
if (solveMazeUtil(maze, x, y + i, sol) == true)
{
return true;
}
}
/* If none of the above movements work then
BACKTRACK: unmark x, y as part of solution
path */
sol[x][y] = 0;
return false;
}
return false;
}
// Driver Code
public static void main(String[] args)
{
int maze[][] = {{2, 1, 0, 0},
{3, 0, 0, 1},
{0, 1, 0, 1},
{0, 0, 0, 1}};
solveMaze(maze);
}
}
// This code is contributed by Princi Singh
Python3
""" Python3 program to solve Rat in a
Maze problem using backtracking """
# Maze size
N = 4
""" A utility function to print solution matrix
sol """
def printSolution(sol):
for i in range(N):
for j in range(N):
print(sol[i][j], end = " ")
print()
""" A utility function to check if
x, y is valid index for N*N maze """
def isSafe(maze, x, y):
# if (x, y outside maze) return false
if (x >= 0 and x < N and y >= 0 and
y < N and maze[x][y] != 0):
return True
return False
""" This function solves the Maze problem using
Backtracking. It mainly uses solveMazeUtil() to
solve the problem. It returns false if no path
is possible, otherwise return True and prints
the path in the form of 1s. Please note that
there may be more than one solutions,
this function prints one of the feasible solutions."""
def solveMaze(maze):
sol = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
if (solveMazeUtil(maze, 0, 0, sol) == False):
print("Solution doesn't exist")
return False
printSolution(sol)
return True
""" A recursive utility function
to solve Maze problem """
def solveMazeUtil(maze, x, y, sol):
# if (x, y is goal) return True
if (x == N - 1 and y == N - 1) :
sol[x][y] = 1
return True
# Check if maze[x][y] is valid
if (isSafe(maze, x, y) == True):
# mark x, y as part of solution path
sol[x][y] = 1
""" Move forward in x direction """
for i in range(1, N):
if (i <= maze[x][y]):
""" Move forward in x direction """
if (solveMazeUtil(maze, x + i,
y, sol) == True):
return True
""" If moving in x direction doesn't give
solution then Move down in y direction """
if (solveMazeUtil(maze, x,
y + i, sol) == True):
return True
""" If none of the above movements work then
BACKTRACK: unmark x, y as part of solution
path """
sol[x][y] = 0
return False
return False
# Driver Code
maze = [[2, 1, 0, 0],
[3, 0, 0, 1],
[0, 1, 0, 1],
[0, 0, 0, 1]]
solveMaze(maze)
# This code is contributed by SHUBHAMSINGH10
C#
// C# program to solve Rat in a Maze problem
// using backtracking
using System;
class GFG
{
// Maze size
static int N = 4;
/* A utility function to print
solution matrix sol[N, N] */
static void printSolution(int [,]sol)
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
Console.Write(" {0} ", sol[i, j]);
}
Console.Write("\n");
}
}
/* A utility function to check if
x, y is valid index for N*N maze */
static Boolean isSafe(int [,]maze,
int x, int y)
{
// if (x, y outside maze) return false
if (x >= 0 && x < N && y >= 0 &&
y < N && maze[x, y] != 0)
{
return true;
}
return false;
}
/* This function solves the Maze problem using
Backtracking. It mainly uses solveMazeUtil() to
solve the problem. It returns false if no path
is possible, otherwise return true and prints
the path in the form of 1s. Please note that
there may be more than one solutions,
this function prints one of the feasible solutions.*/
static Boolean solveMaze(int [,]maze)
{
int [,]sol = {{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0}};
if (solveMazeUtil(maze, 0, 0, sol) == false)
{
Console.Write("Solution doesn't exist");
return false;
}
printSolution(sol);
return true;
}
/* A recursive utility function to solve Maze problem */
static Boolean solveMazeUtil(int [,]maze, int x,
int y, int [,]sol)
{
// if (x, y is goal) return true
if (x == N - 1 && y == N - 1)
{
sol[x, y] = 1;
return true;
}
// Check if maze[x,y] is valid
if (isSafe(maze, x, y) == true)
{
// mark x, y as part of solution path
sol[x, y] = 1;
/* Move forward in x direction */
for (int i = 1;
i <= maze[x, y] && i < N; i++)
{
/* Move forward in x direction */
if (solveMazeUtil(maze, x + i,
y, sol) == true)
{
return true;
}
/* If moving in x direction doesn't give
solution then Move down in y direction */
if (solveMazeUtil(maze, x,
y + i, sol) == true)
{
return true;
}
}
/* If none of the above movements work then
BACKTRACK: unmark x, y as part of solution
path */
sol[x, y] = 0;
return false;
}
return false;
}
// Driver Code
public static void Main(String[] args)
{
int [,]maze = {{2, 1, 0, 0},
{3, 0, 0, 1},
{0, 1, 0, 1},
{0, 0, 0, 1}};
solveMaze(maze);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript program to solve Rat in a Maze problem
// using backtracking
// Maze size
let N = 4;
/* A utility function to print solution matrix
sol[N][N] */
function printSolution(sol)
{
for (let i = 0; i < N; i++)
{
for (let j = 0; j < N; j++)
{
document.write(sol[i][j] + " ");
}
document.write("<br/>");
}
}
/* A utility function to check if x, y is valid
index for N*N maze */
function isSafe(maze, x, y)
{
// if (x, y outside maze) return false
if (x >= 0 && x < N && y >= 0 &&
y < N && maze[x][y] != 0)
{
return true;
}
return false;
}
/* This function solves the Maze problem using
Backtracking. It mainly uses solveMazeUtil() to
solve the problem. It returns false if no path
is possible, otherwise return true and print
the path in the form of 1s. Please note that
there may be more than one solutions,
this function prints one of the feasible solutions.*/
function solveMaze(maze)
{
let sol = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]];
if (solveMazeUtil(maze, 0, 0, sol) == false)
{
document.write("Solution doesn't exist");
return false;
}
printSolution(sol);
return true;
}
/* A recursive utility function to solve Maze problem */
function solveMazeUtil(maze, x,
y, sol)
{
// if (x, y is goal) return true
if (x == N - 1 && y == N - 1)
{
sol[x][y] = 1;
return true;
}
// Check if maze[x][y] is valid
if (isSafe(maze, x, y) == true)
{
// mark x, y as part of solution path
sol[x][y] = 1;
/* Move forward in x direction */
for (let i = 1; i <= maze[x][y] && i < N; i++)
{
/* Move forward in x direction */
if (solveMazeUtil(maze, x + i, y, sol) == true)
{
return true;
}
/* If moving in x direction doesn't give
solution then Move down in y direction */
if (solveMazeUtil(maze, x, y + i, sol) == true)
{
return true;
}
}
/* If none of the above movements work then
BACKTRACK: unmark x, y as part of solution
path */
sol[x][y] = 0;
return false;
}
return false;
}
// Driver code
let maze = [[2, 1, 0, 0],
[3, 0, 0, 1],
[0, 1, 0, 1],
[0, 0, 0, 1]];
solveMaze(maze);
// This code is contributed by splevel62.
</script>
Producción:
1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1
Publicación traducida automáticamente
Artículo escrito por Ajay Verma y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA