一个迷宫被给出为n*n二进制矩阵的块，其中源块是最左上方的块，即Maze[0][0]，目标块是最右下方的块，即Maze[n-1][n-1]。老鼠从源头出发，必须到达目的地。老鼠只能朝两个方向移动：向前和向下。
在迷宫矩阵中，0表示块是死端，1表示块可以用于从源到目标的路径。请注意，这是典型迷宫问题的简单版本。例如，更复杂的版本可能是老鼠可以向四个方向移动，而更复杂的版本可能移动的次数有限。
下面是一个迷宫的例子。
Backtracking 回溯法的三部曲：
1 初始化原始数据，开始点
2 判断下一步是否合法，如果合法就继续递归搜索答案，如果不合法就返回
3 递归直到找到答案，返回真值
这里只需要找到一个解就可以，所以只要找到一个解就可以马上返回。
/*
A Maze is given as N*N binary matrix of blocks where source block is the upper
left most block i.e., maze[0][0] and destination block is lower rightmost
block i.e., maze[N-1][N-1]. A rat starts from source and has to reach destination.
The rat can move only in two directions: forward and down. In the maze matrix,
0 means the block is dead end and 1 means the block can be used in the path
from source to destination.
*/
#include <iostream>
#define size 4
using namespace std;
int solveMaze(int currposrow, int currposcol, int maze[size][size], int soln[size][size])
{
if ((currposrow == size - 1) && (currposcol == size - 1))
{
soln[currposrow][currposcol] = 1;
for (int i = 0; i<size; ++i)
{
for (int j = 0; j<size; ++j)
{
cout << soln[i][j];
}
cout << endl;
}
return 1;
}
else
{
soln[currposrow][currposcol] = 1;
// if there exist a solution by moving one step ahead in a collumn
if ((currposcol<size - 1) && maze[currposrow][currposcol + 1] == 1 && solveMaze(currposrow, currposcol + 1, maze, soln))
{
return 1;
}
// if there exists a solution by moving one step ahead in a row
if ((currposrow<size - 1) && maze[currposrow + 1][currposcol] == 1 && solveMaze(currposrow + 1, currposcol, maze, soln))
{
return 1;
}
// the backtracking part
soln[currposrow][currposcol] = 0;
return 0;
}
}
int main(int argc, char const *argv[])
{
int maze[size][size] = {
{ 1, 0, 1, 0 },
{ 1, 0, 1, 1 },
{ 1, 0, 0, 1 },
{ 1, 1, 1, 1 }
};
int soln[size][size];
for (int i = 0; i<size; ++i)
{
for (int j = 0; j<size; ++j)
{
soln[i][j] = 0;
}
}
int currposrow = 0;
int currposcol = 0;
solveMaze(currposrow, currposcol, maze, soln);
system("pause");
return 0;
}
