这是老鼠在迷宫中的变异
迷宫作为块的N * N二进制矩阵给出，其中源块是最左上的块，即maze [0] [0]，而目标块是最右下块的块，即maze [N-1] [N-1] 。老鼠从源头开始，必须到达目的地。老鼠只能在两个方向上移动：向前和向下。
在迷宫矩阵中，0表示该块是死胡同，非零数字表示该块可以在从源到目标的路径中使用。 mat [i] [j]的非零值表示rat可以从单元mat [i] [j]跳出的最大跳数。
在此变体中，Rat允许一次跳多个步骤，而不是1。
例子：
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天真的算法
天真的算法是生成从源到目的地的所有路径，并一一检查生成的路径是否满足约束条件。
while there are untried paths
{
generate the next path
if this path has all blocks as non-zero
{
print this path;
}
}回溯算法
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.回溯解决方案的实施
C++
/* C/C++ program to solve Rat in a Maze problem
using backtracking */
#include
// 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 SinghPython3
""" Python3 program to solve Rat in a
Maze problem using backtracking """
# Maze size
N = 4
""" A utility function to prsolution 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 SHUBHAMSINGH10C#
// 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 29AjayKumarJavascript
输出：
1 0 0 0
1 0 0 1
0 0 0 1
0 0 0 1