迷宫中的老鼠问题是众所周知的回溯问题之一。在这里我们将看到这个问题几乎没有变化。假设给定一个NxN迷宫M。起点为左上角 M[0, 0]，终点为右下角 M[N – 1, N - 1]。一只老鼠被放置在起点。我们的目标是找到一条从起点到终点的路径，可以让老鼠到达目的地。这里老鼠可以跳跃（变体）。现在有一些限制
- 老鼠可以向右或向下移动。
- 迷宫中单元格中带有 0 表示该单元格是被阻止。
- 非零单元格表示有效路径。
- 单元格内的数字表示大鼠从该单元格可以进行的最大跳跃次数。 ul>
算法
ratInMaze
begin
if destination is reached, then
print the solution matrix
else
1. Place the current cell inside the solution matrix as 1
2. Move forward or jump (check max jump value) and recursively check if move leads to solution or not.
3. If the move taken from the step 2 is not correct, then move down, and check it leads to the solution or not
4. If none of the solutions in step 2 and 3 are correct, then make the current cell 0.
end if
end示例
#include <iostream>
#define N 4
using namespace std;
void dispSolution(int sol[N][N]) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
cout << sol[i][j] << " ";
cout << endl;
}
}
bool isSafe(int maze[N][N], int x, int y) { //check whether x,y is valid or not
// when (x, y) is outside of the maze, then return false
if (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] != 0)
return true;
return false;
}
bool ratMazeSolve(int maze[N][N], int x, int y, int sol[N][N]) {
if (x == N - 1 && y == N - 1) { //if destination is found, return true
sol[x][y] = 1;
return true;
}
if (isSafe(maze, x, y)) {
sol[x][y] = 1; //mark 1 into solution matrix
for (int i = 1; i <= maze[x][y] && i < N; i++) {
if (ratMazeSolve(maze, x + i, y, sol)) //move right
return true;
if (ratMazeSolve(maze, x, y + i, sol)) //move down
return true;
}
sol[x][y] = 0; //if the solution is not valid, then make it 0
return false;
}
return false;
}
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 (!ratMazeSolve(maze, 0, 0, sol)) {
cout << "Solution doesn't exist";
return false;
}
dispSolution(sol);
return true;
}
main() {
int maze[N][N] = { { 2, 1, 0, 0 },
{ 3, 0, 0, 1 },
{ 0, 1, 0, 1 },
{ 0, 0, 0, 1 }
};
solveMaze(maze);
}输出
1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1
