C语言使用深度优先搜索算法解决迷宫问题(堆栈)

将起点标记为已走过并压栈;while (栈非空) {从栈顶弹出一个点p;if (p这个点是终点)break;否则沿右、下、左、上四个方向探索相邻的点if (和p相邻的点有路可走，并且还没走过)将相邻的点标记为已走过并压栈，它的前趋就是p点;}if (p点是终点) {打印p点的坐标;while (p点有前趋) {p点 = p点的前趋;打印p点的坐标;}} else没有路线可以到达终点;

#include <stdio.h>#define MAX_ROW 5#define MAX_COL 5struct point { int row, col; } stack[512];int top = 0;void push(struct point p){  stack[top++] = p;}struct point pop(void){  return stack[--top];}int is_empty(void){  return top == 0;}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");}struct point predecessor[MAX_ROW][MAX_COL] = {  {{-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}},  {{-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}},  {{-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}},  {{-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}},  {{-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}},};void visit(int row, int col, struct point pre){  struct point visit_point = { row, col };  maze[row][col] = 2;  predecessor[row][col] = pre;  push(visit_point);}int main(void){  struct point p = { 0, 0 };  maze[p.row][p.col] = 2;  push(p);  while (!is_empty()) {    p = pop();    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, p);    if (p.row+1 < MAX_ROW /* down */ && maze[p.row+1][p.col] == 0)      visit(p.row+1, p.col, p);    if (p.col-1 >= 0 /* left */ && maze[p.row][p.col-1] == 0)      visit(p.row, p.col-1, p);    if (p.row-1 >= 0 /* up */  && maze[p.row-1][p.col] == 0)      visit(p.row-1, p.col, p);    print_maze();  }  if (p.row == MAX_ROW - 1 && p.col == MAX_COL - 1)  {    printf("(%d, %d)/n", p.row, p.col);    while (predecessor[p.row][p.col].row != -1) {      p = predecessor[p.row][p.col];      printf("(%d, %d)/n", p.row,      p.col);    }  } else    printf("No path!/n");    return 0;}