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 19:03:17 浏览: 20
以下是用C语言解决该问题的代码实现:
```c
#include <stdio.h>
#include <string.h>
#define MAX_N 6
#define MAX_M 6
char maze[MAX_N][MAX_M];
int visited[MAX_N][MAX_M];
int start_x, start_y, end_x, end_y;
int num_of_ways = 0;
void dfs(int x, int y) {
if (x == end_x && y == end_y) { // 到达终点,记录一种方案
num_of_ways++;
return;
}
visited[x][y] = 1;
if (x - 1 >= 0 && maze[x-1][y] == 'T' && visited[x-1][y] == 0) { // 向上移动
dfs(x-1, y);
}
if (x + 1 < MAX_N && maze[x+1][y] == 'T' && visited[x+1][y] == 0) { // 向下移动
dfs(x+1, y);
}
if (y - 1 >= 0 && maze[x][y-1] == 'T' && visited[x][y-1] == 0) { // 向左移动
dfs(x, y-1);
}
if (y + 1 < MAX_M && maze[x][y+1] == 'T' && visited[x][y+1] == 0) { // 向右移动
dfs(x, y+1);
}
visited[x][y] = 0; // 回溯
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%s", maze[i]);
for (int j = 0; j < m; j++) {
if (maze[i][j] == 'B') { // 记录起点位置
start_x = i;
start_y = j;
} else if (maze[i][j] == 'E') { // 记录终点位置
end_x = i;
end_y = j;
}
}
}
dfs(start_x, start_y);
printf("%d\n", num_of_ways);
return 0;
}
```
该代码使用深度优先搜索算法来解决问题。具体地,从起点开始,依次向四个方向进行探索,如果可以移动到相邻格子,就继续往前搜索。同时,为了避免重复经过起点和终点,我们需要记录哪些格子已经被访问过,每次回溯时需要将其重置。最终,我们统计可以到达终点的方案数即可。