迷宫问题（一）-利用栈回溯

#include<stdio.h>
#include <stdlib.h>
#define OK      1
#define ERROR   0
#define TRUE    1
#define FALSE   0
#define OVERFLOW -1
#define STACK_INIT_SIZE 100
#define STACKINCREMENT 20
#define N 10

typedef int Status;
typedef char MazeType[N][N + 1];

int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};

typedef struct {
  int x; //迷宫中的行
  int y; //迷宫中的列
} PosType; //坐标

typedef struct {
  PosType      seat;    //当前的坐标位置
  int   di;        //往下一个坐标位置的方向
} SElemType; //栈的元素类型

typedef struct {
  SElemType *base;
  SElemType *top;
  int stacksize;
} SqStack;

Status InitStack(SqStack &s)  //栈的初始化
{
  s.base = (SElemType *) malloc(STACK_INIT_SIZE * sizeof(SElemType));
  if(!s.base) exit(OVERFLOW);
  s.top = s.base;
  s.stacksize = STACK_INIT_SIZE;
  return OK;
}

Status Push(SqStack &s, SElemType e)
{
  if(s.top - s.base >= s.stacksize) { //栈满，追加存储空间
    s.base = (SElemType *)realloc(s.base, (s.stacksize + STACKINCREMENT) * sizeof(SElemType));
    if(!s.base) exit(OVERFLOW);
    s.top = s.base + s.stacksize;
    s.stacksize += STACKINCREMENT;
  }
  *s.top++ = e;
  return OK;
}

Status Pop(SqStack &s, SElemType &e)
{
  if(s.top == s.base)return ERROR;
  e = * --s.top;
  return OK;
}

int StackEmpty(SqStack s)
{
  return s.base == s.top;
}

Status Pass(MazeType maze, PosType curpos) //判断当前节点是否通过
{
  return maze[curpos.x][curpos.y] == ‘ ‘ || maze[curpos.x][curpos.y] == ‘S‘
  || maze[curpos.x][curpos.y] == ‘E‘;
}

void FootPrint(MazeType &maze, PosType curpos) //留下足迹
{
  maze[curpos.x][curpos.y] = ‘*‘; //走过且走得通
}

PosType NextPos(PosType pos,int d){
    pos.x += dx[d];
    pos.y += dy[d];
    return pos;
}

void MarkPrint(MazeType &maze, PosType curpos) //留下不能通过的标记
{
  maze[curpos.x][curpos.y] = ‘!‘; //走过但走不通
}

//求解迷宫maze中,从入口start到出口end的一条路径
Status MazePath(MazeType &maze, PosType start, PosType end)
{
  SqStack s;
  SElemType e;
  InitStack(s);
  PosType curpos = start;
  do {
    if(Pass(maze, curpos)) { //如果当前位置可以通过,即是未曾走到的通道块
      FootPrint(maze, curpos);     //留下足迹
      if(curpos.x==end.x && curpos.y==end.y)
        return TRUE;
      e.seat = curpos;
      e.di = 0;
      Push(s, e);
      curpos = NextPos(curpos, 0);  //获得下一节点:当前位置的东邻
    }
    else { //当前位置不能通过
      if(!StackEmpty(s)) {
        Pop(s, e);
        while(e.di == 3 && !StackEmpty(s)) {
          MarkPrint(maze, e.seat);
          Pop(s, e);  //留下不能通过的标记,并退回一步
        }
        if(e.di < 3) {
          e.di++;
          Push(s, e);           //换一个方向探索
          curpos = NextPos(e.seat, e.di);   //设定当前位置是该方向上的相邻块
        }//if
      }//if
    }//else
  } while(!StackEmpty(s));
  return FALSE;
}
int main()
{
  int i;
  PosType start={1, 1}, end={8,8};
  MazeType maze;
  for(i = 0; i < 10; i++)
    gets(maze[i]);
  MazePath(maze, start, end);
  for(i = 0; i < 10; i++)
    puts(maze[i]);
  return 0;
}