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c - 洪水填充算法迷宫

转载 作者:行者123 更新时间:2023-11-30 20:00:59 26 4
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我试图让我的程序读取这样的迷宫:

    #.#######
#.......#
####.####
#....#..#
#.####.##

并打印出迷宫中的可达区域和不可到达区域,应该如下所示:

    #+#######
#+++++++#
####+####
#++++#--#
#+####-##

墙壁用“#”表示,可通过的单元格用“.”表示。

替换单元格的“+”表示可以从迷宫的顶部入口点到达这些单元格。 “-”符号是进入迷宫顶部无法到达的单元格。

例如,在上面的迷宫中,除了右下角的单元格之外,所有单元格都是可到达的。这是因为无法从迷宫顶部的入口点到达这些单元格。

我正在尝试使用一些递归来淹没迷宫,并确定可到达的区域,但我遇到了麻烦。

这是我到目前为止所拥有的:

    int
flood_fill(m_t * maze, int row, int col) {

int direction;

direction = flood_fill(maze, row+1, col); /* down */
if (!direction) {
direction = flood_fill(maze, row, col+1); /* right */
}

if (!direction) {
direction = flood_fill(maze, row-1, col); /* up */
}

if (!direction) {
direction = flood_fill(maze, row, col-1); /* left */
}

if (direction) {
maze->M[row][col].type = path;
}

return direction;
}

我知道我的flood_fill函数没有做正确的事情,而且我很难做到正确。任何人都可以帮助我,让我的洪水填充部分代码正确,这样我就可以在代码中的其他地方调用该函数并确定可以到达哪些单元格。

最佳答案

试试这个

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define MAX_HEIGHT 100
#define MAX_WIDTH 100
#define wall '#'
#define path_cell '.'

typedef struct {
char type;
int reachable;
int visited;
} mazecells_t;

typedef struct {
int height;
int width;
mazecells_t M[MAX_HEIGHT][MAX_WIDTH];
} m_t;

void readmaze(m_t *maze);
void print(m_t *m);
void print_reachable(m_t *m);

int main(void) {
m_t MAZE;

readmaze(&MAZE);
print(&MAZE);
puts("");
print_reachable(&MAZE);

return 0;
}

void readmaze(m_t *maze) {
int row=0, col=0;
char ch;
FILE *fp = stdin;//fopen("map.txt", "r");
while(EOF != fscanf(fp, "%c", &ch)){
if(ch == wall || ch == path_cell){
maze->M[row][col].type = ch;
maze->M[row][col].reachable = 0;
maze->M[row][col].visited = 0;
++col;
} else if(ch == '\n'){
maze->width = col;
col = 0;
++row;
}
}
if(col != 0)
++row;
//fclose(fp);
maze->height = row;
}

void print(m_t *m){
for(int r = 0; r < m->height; ++r){
for(int c = 0; c < m->width; ++c){
putchar(m->M[r][c].type);
}
putchar('\n');
}
}

typedef enum dir {
UP, RIGHT, DOWN, LEFT, FORWARD
} DIR;

typedef struct pos {
int r, c;
DIR dir;
} Pos;

typedef struct node {
Pos pos;
struct node *next;
} Node;

typedef struct queque {
Node *head, *tail;
} Queque;

Queque *new_queque(void){
Queque *q = malloc(sizeof(*q));
q->head = q->tail = NULL;
return q;
}

void enque(Queque *q, Pos pos){
Node *node = malloc(sizeof(*node));
node->pos = pos;
node->next = NULL;
if(q->head == NULL){
q->head = q->tail = node;
} else {
q->tail = q->tail->next = node;
}
}

Pos deque(Queque *q){
Pos pos = q->head->pos;
Node *temp = q->head;
if((q->head = q->head->next)==NULL)
q->tail = NULL;
free(temp);
return pos;
}

bool isEmpty_que(Queque *q){
return q->head == NULL;
}

Pos dxdy(DIR curr, DIR next){
Pos d = { 0, 0, 0};
switch(curr){
case UP:
switch(next){
case LEFT:
d.c -= 1;
break;
case FORWARD:
d.r -= 1;
break;
case RIGHT:
d.c += 1;
break;
}
break;
case RIGHT:
switch(next){
case LEFT:
d.r -= 1;
break;
case FORWARD:
d.c += 1;
break;
case RIGHT:
d.r += 1;
break;
}
break;
case DOWN:
switch(next){
case LEFT:
d.c += 1;
break;
case FORWARD:
d.r += 1;
break;
case RIGHT:
d.c -= 1;
break;
}
break;
case LEFT:
switch(next){
case LEFT:
d.r += 1;
break;
case FORWARD:
d.c -= 1;
break;
case RIGHT:
d.r -= 1;
break;
}
break;
}
return d;
}

Pos next_pos(Pos pos, DIR dir){
Pos dxy = dxdy(pos.dir, dir);
switch(dir){
case RIGHT:
pos.dir = (pos.dir + 1) % 4;
break;
case LEFT:
if((pos.dir = (pos.dir - 1)) < 0)
pos.dir += 4;
break;
case FORWARD:
break;
}
pos.r += dxy.r;
pos.c += dxy.c;
return pos;
}
static inline bool isValid(m_t *m, Pos pos){
if(pos.r < 0 || pos.r >= m->height || pos.c < 0 || pos.c >= m->width || m->M[pos.r][pos.c].type == wall)
return false;
return true;
}
static inline bool isValidAndUnvisit(m_t *m, Pos pos){
return isValid(m, pos) && !m->M[pos.r][pos.c].reachable;
}

void print_reachable(m_t *m){
int i;
for(i = 0; i < m->width; ++i)
if(m->M[0][i].type == path_cell)
break;
Pos pos = { 0, i, DOWN};
Queque *q = new_queque();
enque(q, pos);
while(!isEmpty_que(q)){
pos = deque(q);
if(!m->M[pos.r][pos.c].reachable){
m->M[pos.r][pos.c].reachable = 1;

Pos next = next_pos(pos, LEFT);
if(isValidAndUnvisit(m, next))
enque(q, next);
next = next_pos(pos, FORWARD);
if(isValidAndUnvisit(m, next))
enque(q, next);
next = next_pos(pos, RIGHT);
if(isValidAndUnvisit(m, next))
enque(q, next);
}
}
free(q);
for(int r = 0; r < m->height; ++r){
for(int c = 0; c < m->width; ++c){
if(m->M[r][c].reachable)
putchar('+');
else if(m->M[r][c].type == path_cell)
putchar('-');
else
putchar(m->M[r][c].type);
}
putchar('\n');
}
}

关于c - 洪水填充算法迷宫,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37207022/

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