用c语言解决下述问题:描述: Erik is a knight who can save the princess out of the maze, so Erik walked into the maze, a 4-direction maze (left, up, right, down), and found the princess. Now Erik wants to walk to the exit of the maze. Luckily Erik has a map of this maze. The maze consists of n*m grids, with each grid being one of the following types: T: representing Erik can walk to this grid. F: representing Erik cannot walk to this grid, because of danger. B: representing Erik's beginning position. E: representing the exit of the maze. If the grid's type is T and Erik walks from this gird to another, then the gird's type will change to F. Also, Erik cannot pass B and E twice or more. Now, Erik wants to know how many different ways he can get out of the maze. 输入:The first line contains two integers n and m (1<=n<=6,1<=m<=6). There are n more lines in the case, each of which includes m characters. Each character represents the type of a grid, that is, T, F, B or E. 输出:Output a number representing the number of different ways in which Erik can get out of the maze.
时间: 2024-02-19 10:57:23 浏览: 25
以下是用C语言实现该问题的代码:
```c
#include <stdio.h>
int n, m, count = 0;
char maze[6][6];
void dfs(int x, int y) {
if (maze[x][y] == 'E') { // 到达出口
count++;
return;
}
if (maze[x][y] == 'F' || maze[x][y] == 'B') { // 不可到达或已经访问过
return;
}
char temp = maze[x][y];
maze[x][y] = 'B'; // 标记当前位置为已访问
dfs(x-1, y); // 上
dfs(x+1, y); // 下
dfs(x, y-1); // 左
dfs(x, y+1); // 右
maze[x][y] = temp; // 恢复当前位置的状态
}
int main() {
scanf("%d %d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%s", maze[i]);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (maze[i][j] == 'B') { // 从起点开始遍历
dfs(i, j);
break;
}
}
}
printf("%d\n", count);
return 0;
}
```
该代码使用深度优先搜索(DFS)算法解决该问题,定义了一个 `dfs()` 函数,用于递归遍历迷宫,并使用一个全局变量 `count` 记录可行的路径数。首先读入迷宫的大小和每个位置的类型,然后从起点开始遍历,每次遍历到一个可达的位置时将其标记为已访问,并向上、下、左、右四个方向递归遍历,最后恢复当前位置的状态。另外,如果遍历到终点,将 `count` 加 1。最后输出 `count` 即可得到可行的路径数。