memset(maze.mazeMap, false, sizeof(maze.mazeMap));
时间: 2023-08-18 17:10:44 浏览: 44
这段代码使用了C++标准库头文件`cstring`中的`memset`函数,用于对`maze.mazeMap`数组中的所有元素进行初始化为`false`。
具体来说,`memset`函数的语法如下:
```c++
void *memset( void *dest, int ch, std::size_t count );
```
其中,`dest`为要初始化的内存地址,`ch`为初始化内容,`count`为初始化的字节数。
在上述代码中,`maze.mazeMap`为要初始化的内存地址,`false`为初始化内容,`sizeof(maze.mazeMap)`为初始化的字节数,即`maze.mazeMap`数组中元素的数量乘上每个元素所占用的字节数。
因此,这段代码的作用是将`maze.mazeMap`数组中的所有元素初始化为`false`。
相关问题
用c语言解决这个问题: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.
这个问题可以使用深度优先搜索(DFS)算法来解决。具体步骤如下:
1. 读入输入数据,将起点和终点的位置记录下来。
2. 定义一个二维数组 maze 存储迷宫的地图,使用一个一维数组 visited 记录每个位置是否已经被访问过。
3. 定义一个计数器 count,初始值为 0。
4. 定义一个 DFS 函数,输入参数为当前位置的行、列坐标和已经走过的步数 step。
5. 如果当前位置已经是终点位置,则将 count 加 1。
6. 否则,遍历当前位置的四个邻居位置(上下左右),如果该邻居位置是合法的、未被访问过且可以到达(即该位置是 T),则将该邻居位置的 visited 标记为已访问,并将该位置的类型改为 F,然后递归调用 DFS 函数,更新当前位置为邻居位置,步数加一,并将新的位置和步数作为参数传给 DFS 函数。
7. 在 DFS 函数结束后,将当前位置的 visited 标记为未访问过,将该位置的类型改回 T(回溯过程)。
8. 最后输出 count 的值即可。
以下是完整的 C 语言代码实现:
```c
#include <stdio.h>
#include <string.h>
#define MAX_N 6
#define MAX_M 6
int n, m, start_row, start_col, end_row, end_col;
char maze[MAX_N][MAX_M + 1];
int visited[MAX_N][MAX_M];
int count = 0;
void dfs(int row, int col, int step)
{
if (row == end_row && col == end_col) {
count++;
return;
}
visited[row][col] = 1;
int dr[4] = {-1, 0, 1, 0};
int dc[4] = {0, 1, 0, -1};
for (int i = 0; i < 4; i++) {
int new_row = row + dr[i];
int new_col = col + dc[i];
if (new_row >= 0 && new_row < n && new_col >= 0 && new_col < m
&& !visited[new_row][new_col] && maze[new_row][new_col] == 'T') {
maze[new_row][new_col] = 'F';
dfs(new_row, new_col, step + 1);
maze[new_row][new_col] = 'T';
}
}
visited[row][col] = 0;
}
int main()
{
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') {
start_row = i;
start_col = j;
} else if (maze[i][j] == 'E') {
end_row = i;
end_col = j;
}
}
}
memset(visited, 0, sizeof(visited));
dfs(start_row, start_col, 0);
printf("%d\n", count);
return 0;
}
```
注意,本题中 n 和 m 的范围比较小,因此可以使用递归的深度优先搜索算法。如果 n 和 m 的范围较大,可能需要使用迭代的深度优先搜索算法或广度优先搜索算法来解决。
#include <iostream> #include <stack> #include <cstdlib> #include <ctime> using namespace std; const int MAXN = 100; const char WALL = '#'; const char PATH = ' '; const char START = 'S'; const char END = 'E'; const int dx[4] = { -1, 0, 1, 0 }; const int dy[4] = { 0, 1, 0, -1 }; int n, m; char maze[MAXN][MAXN]; bool vis[MAXN][MAXN]; stack<pair<int, int>> st; void init() { // 随机生成迷宫 srand(time(NULL)); n = rand() % 10 + 5; m = rand() % 10 + 5; int sx = rand() % n; int sy = rand() % m; int ex = rand() % n; int ey = rand() % m; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (i == sx && j == sy) maze[i][j] = START; else if (i == ex && j == ey) maze[i][j] = END; else if (rand() % 4 == 0) maze[i][j] = WALL; else maze[i][j] = PATH; } } } void print() { // 输出迷宫 cout << "Maze:" << endl; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { cout << maze[i][j] << ' '; } cout << endl; } } bool dfs(int x, int y) { // 深度优先搜索 vis[x][y] = true; st.push(make_pair(x, y)); if (maze[x][y] == END) { return true; } for (int i = 0; i < 4; i++) { int nx = x + dx[i]; int ny = y + dy[i]; if (nx >= 0 && nx < n && ny >= 0 && ny < m && maze[nx][ny] != WALL && !vis[nx][ny]) { if (dfs(nx, ny)) { return true; } } } st.pop(); return false; } void solve() { // 求解迷宫 memset(vis, false, sizeof(vis)); while (!st.empty()) st.pop(); dfs(0, 0); } void print_path() { // 输出路径 cout << "Path:" << endl; while (!st.empty()) { auto p = st.top(); st.pop(); cout << '(' << p.first << ", " << p.second << ')' << endl; } } int main() { init(); print(); solve(); print_path(); return 0; } 在这个代码的基础上,添加路径不存在的情况
在代码中,我们可以在dfs函数中添加一个判断,如果搜索完所有可行的路径后,仍然没有找到终点,那么就说明路径不存在,此时返回false即可。
具体来说,我们可以在dfs函数的末尾添加一个判断:
```c++
if (st.empty()) {
cout << "Path does not exist" << endl;
return false;
}
```
完整代码如下:
```c++
#include <iostream>
#include <stack>
#include <cstdlib>
#include <ctime>
using namespace std;
const int MAXN = 100;
const char WALL = '#';
const char PATH = ' ';
const char START = 'S';
const char END = 'E';
const int dx[4] = { -1, 0, 1, 0 };
const int dy[4] = { 0, 1, 0, -1 };
int n, m;
char maze[MAXN][MAXN];
bool vis[MAXN][MAXN];
stack<pair<int, int>> st;
void init() {
// 随机生成迷宫
srand(time(NULL));
n = rand() % 10 + 5;
m = rand() % 10 + 5;
int sx = rand() % n;
int sy = rand() % m;
int ex = rand() % n;
int ey = rand() % m;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == sx && j == sy) maze[i][j] = START;
else if (i == ex && j == ey) maze[i][j] = END;
else if (rand() % 4 == 0) maze[i][j] = WALL;
else maze[i][j] = PATH;
}
}
}
void print() {
// 输出迷宫
cout << "Maze:" << endl;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cout << maze[i][j] << ' ';
}
cout << endl;
}
}
bool dfs(int x, int y) {
// 深度优先搜索
vis[x][y] = true;
st.push(make_pair(x, y));
if (maze[x][y] == END) {
return true;
}
for (int i = 0; i < 4; i++) {
int nx = x + dx[i];
int ny = y + dy[i];
if (nx >= 0 && nx < n && ny >= 0 && ny < m && maze[nx][ny] != WALL && !vis[nx][ny]) {
if (dfs(nx, ny)) {
return true;
}
}
}
st.pop();
if (st.empty()) {
cout << "Path does not exist" << endl;
return false;
}
}
void solve() {
// 求解迷宫
memset(vis, false, sizeof(vis));
while (!st.empty()) st.pop();
dfs(0, 0);
}
void print_path() {
// 输出路径
cout << "Path:" << endl;
while (!st.empty()) {
auto p = st.top();
st.pop();
cout << '(' << p.first << ", " << p.second << ')' << endl;
}
}
int main() {
init();
print();
solve();
print_path();
return 0;
}
```