分别用广度优先和深度优先解决迷宫问题c语言
时间: 2023-07-31 22:08:02 浏览: 123
好的,下面是分别用广度优先和深度优先解决迷宫问题的 C 语言代码,供参考:
### 广度优先搜索
```c
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#define ROW 5
#define COL 5
int maze[ROW][COL] = {
{0, 1, 0, 0, 0},
{0, 1, 0, 1, 0},
{0, 0, 0, 0, 0},
{0, 1, 1, 1, 0},
{0, 0, 0, 1, 0}
};
bool visited[ROW][COL];
typedef struct {
int x;
int y;
int prev;
} Point;
Point queue[ROW * COL];
int front = -1, rear = -1;
void enqueue(Point p) {
if (front == -1) {
front++;
}
queue[++rear] = p;
}
Point dequeue() {
if (front > rear) {
printf("Queue is empty!");
exit(1);
}
return queue[front++];
}
bool is_valid(int x, int y) {
if (x < 0 || x >= ROW || y < 0 || y >= COL) {
return false;
}
if (maze[x][y] == 1 || visited[x][y]) {
return false;
}
return true;
}
void print_path(Point p) {
if (p.prev == -1) {
printf("(%d, %d)", p.x, p.y);
return;
}
print_path(queue[p.prev]);
printf(" -> (%d, %d)", p.x, p.y);
}
void bfs(int x, int y) {
visited[x][y] = true;
Point start = {x, y, -1};
enqueue(start);
while (front <= rear) {
Point curr = dequeue();
if (curr.x == ROW - 1 && curr.y == COL - 1) {
print_path(curr);
return;
}
int x_dir[] = {-1, 0, 1, 0};
int y_dir[] = {0, 1, 0, -1};
for (int i = 0; i < 4; i++) {
int next_x = curr.x + x_dir[i];
int next_y = curr.y + y_dir[i];
if (is_valid(next_x, next_y)) {
visited[next_x][next_y] = true;
Point next = {next_x, next_y, front - 1};
enqueue(next);
}
}
}
printf("No path found!");
}
int main() {
bfs(0, 0);
return 0;
}
```
### 深度优先搜索
```c
#include <stdio.h>
#include <stdbool.h>
#define ROW 5
#define COL 5
int maze[ROW][COL] = {
{0, 1, 0, 0, 0},
{0, 1, 0, 1, 0},
{0, 0, 0, 0, 0},
{0, 1, 1, 1, 0},
{0, 0, 0, 1, 0}
};
bool visited[ROW][COL];
bool dfs(int x, int y) {
if (x < 0 || x >= ROW || y < 0 || y >= COL) {
return false;
}
if (maze[x][y] == 1 || visited[x][y]) {
return false;
}
visited[x][y] = true;
if (x == ROW - 1 && y == COL - 1) {
printf("(%d, %d)", x, y);
return true;
}
if (dfs(x - 1, y) || dfs(x, y + 1) || dfs(x + 1, y) || dfs(x, y - 1)) {
printf(" -> (%d, %d)", x, y);
return true;
}
return false;
}
int main() {
if (!dfs(0, 0)) {
printf("No path found!");
}
return 0;
}
```
以上两段代码都是基于同样的 5x5 迷宫,其中 0 表示可以通过,1 表示不能通过。广度优先搜索使用了队列,在每次取出队首元素时,将其合法的相邻点加入队列,并标记为已访问。深度优先搜索使用递归的方式,依次搜索当前点的四周点,如果其中有一条路能到达终点,则输出路径。
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