C语言使用广度优先搜索算法解决迷宫问题（队列）

#include <stdio.h>
#define MAX_ROW 5
#define MAX_COL 5
struct point { int row, col, predecessor; } queue[512];
int head = 0, tail = 0;
void enqueue(struct point p)
{
  queue[tail++] = p;
}
struct point dequeue(void)
{
  return queue[head++];
}
int is_empty(void)
{
  return head == tail;
}
int maze[MAX_ROW][MAX_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,
};
void print_maze(void)
{
  int i, j;
  for (i = 0; i < MAX_ROW; i++) {
    for (j = 0; j < MAX_COL; j++)
    printf("%d ", maze[i][j]);
    putchar('\n');
  }
  printf("*********\n");
}
void visit(int row, int col)
{
  struct point visit_point = { row, col, head-1 };
  maze[row][col] = 2;
  enqueue(visit_point);
}
int main(void)
{
  struct point p = { 0, 0, -1 };
  maze[p.row][p.col] = 2;
  enqueue(p);
  while (!is_empty()) {
    p = dequeue();
    if (p.row == MAX_ROW - 1 /* goal */
    && p.col == MAX_COL - 1)
    break;
    if (p.col+1 < MAX_COL /* right */
    && maze[p.row][p.col+1] == 0)
    visit(p.row, p.col+1);
    if (p.row+1 < MAX_ROW /* down */
    && maze[p.row+1][p.col] == 0)
    visit(p.row+1, p.col);
    if (p.col-1 >= 0 /* left */
    && maze[p.row][p.col-1] == 0)
    visit(p.row, p.col-1);
    if (p.row-1 >= 0 /* up */
    && maze[p.row-1][p.col] == 0)
    visit(p.row-1, p.col);
    print_maze();
  }
  if (p.row == MAX_ROW - 1 && p.col == MAX_COL - 1)
  {
    printf("(%d, %d)\n", p.row, p.col);
    while (p.predecessor != -1) {
      p = queue[p.predecessor];
      printf("(%d, %d)\n", p.row, p.col);
    }
  } else
  printf("No path!\n");
  return 0;
}

[root@localhost arithmetic]# ./maze2.out
2 1 0 0 0
2 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0
*********
2 1 0 0 0
2 1 0 1 0
2 0 0 0 0
0 1 1 1 0
0 0 0 1 0
*********
2 1 0 0 0
2 1 0 1 0
2 2 0 0 0
2 1 1 1 0
0 0 0 1 0
*********
2 1 0 0 0
2 1 0 1 0
2 2 2 0 0
2 1 1 1 0
0 0 0 1 0
*********
2 1 0 0 0
2 1 0 1 0
2 2 2 0 0
2 1 1 1 0
2 0 0 1 0
*********
2 1 0 0 0
2 1 2 1 0
2 2 2 2 0
2 1 1 1 0
2 0 0 1 0
*********
2 1 0 0 0
2 1 2 1 0
2 2 2 2 0
2 1 1 1 0
2 2 0 1 0
*********
2 1 0 0 0
2 1 2 1 0
2 2 2 2 2
2 1 1 1 0
2 2 0 1 0
*********
2 1 2 0 0
2 1 2 1 0
2 2 2 2 2
2 1 1 1 0
2 2 0 1 0
*********
2 1 2 0 0
2 1 2 1 0
2 2 2 2 2
2 1 1 1 0
2 2 2 1 0
*********
2 1 2 0 0
2 1 2 1 2
2 2 2 2 2
2 1 1 1 2
2 2 2 1 0
*********
2 1 2 2 0
2 1 2 1 2
2 2 2 2 2
2 1 1 1 2
2 2 2 1 0
*********
2 1 2 2 0
2 1 2 1 2
2 2 2 2 2
2 1 1 1 2
2 2 2 1 0
*********
2 1 2 2 0
2 1 2 1 2
2 2 2 2 2
2 1 1 1 2
2 2 2 1 2
*********
2 1 2 2 2
2 1 2 1 2
2 2 2 2 2
2 1 1 1 2
2 2 2 1 2
*********
2 1 2 2 2
2 1 2 1 2
2 2 2 2 2
2 1 1 1 2
2 2 2 1 2
*********
(4, 4)
(3, 4)
(2, 4)
(2, 3)
(2, 2)
(2, 1)
(2, 0)
(1, 0)
(0, 0)