迷宫问题

最新推荐文章于 2025-02-25 21:39:05 发布

LeleLJ

最新推荐文章于 2025-02-25 21:39:05 发布

阅读量610

点赞数

CC 4.0 BY-SA版权

分类专栏： C/C++

本文链接：https://blog.youkuaiyun.com/u012085815/article/details/11616267

C/C++ 专栏收录该内容

15 篇文章

订阅专栏

点击(此处)折叠或打开

/*
*题二：求解迷宫的一条最短路径，要求：用栈和队列实现，
*不允许使用递归算法。问题描述见教材第50页“3.2.4 迷宫求解”。
*/
#include <stdio.h>
#include <malloc.h>
#include <conio.h>
#define ERROR 0
#define OK 1
#define OVERFLOW 0
#define STACK_INIT_SIZE 100
#define STACKINCREMENT 100
typedef struct {
int x;
int y;
}PosType;
typedef struct {
int ord;
PosType seat;
int di;
}ElemType;
typedef struct {
ElemType *base;
ElemType *top;
int stacksize;
}SqStack;
int InitStack (SqStack *S)
{
S->base=(ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));
if (!S->base)
exit (OVERFLOW);
S->top=S->base;
S->stacksize=STACK_INIT_SIZE;
return OK;
}
int GetTop (SqStack S, ElemType *e)
{
if (S.top==S.base)
return ERROR;
*e=*(S.top-1);
return OK;
}
int Push (SqStack *S, ElemType e)
{
if (S->top - S->base >= S->stacksize)
{
S->base=(ElemType *)realloc(S->base,
(S->stacksize + STACKINCREMENT)* sizeof(ElemType));
if (!S->base)
exit (OVERFLOW);
S->top=S->base+S->stacksize;
S->stacksize+=STACKINCREMENT;
}
*S->top++ = e;
return OK;
}
int Pop(SqStack *S,ElemType *e)
{
if (S->top==S->base)
return ERROR;
*e=*(--S->top);
return OK;
}
int StackEmpty (SqStack S)
{
if (S.base==S.top)
return 1;
else
return 0;
}
void ShowAll (SqStack *S)
{
ElemType t;
printf ("h l step\n");
while (1)
{
if (Pop (S, &t))
{
printf ("%d %d %d\n", t.seat.x, t.seat.y, t.ord);
}
else
break;
}
}
void Change (SqStack *From, SqStack *To) /* 逆置矩阵 */
{
ElemType t;
while (1)
{
if (Pop (From, &t))
Push (To, t);
else
break;
}
}
int Pass(int maze[10][10], PosType local) /*当前位置是否可走*/
{
if (maze[local.x][local.y]==1)
return 0;
else
return 1;
}
void FootPrint (int maze[10][10],PosType local)/*修改当前位置为1*/
{
maze[local.x][local.y]=1;
}
PosType NextPos (PosType local ,int i)
{
if (i==1)
{
local.y+=1;
}
else if (i==2)
{
local.x+=1;
}
else if (i==3)
{
local.x-=1;
}
return local;
}
void MarkPrint (int maze[10][10],PosType local)/*修改当前位置为1*/
{
maze[local.x][local.y]=1;
}
int MazePath (int maze[10][10],PosType start,PosType end,SqStack *S)
{
PosType local = start;
ElemType temp;
int step=1;
do {
if (Pass (maze, local))
{
FootPrint (maze, local);
temp.ord=step;
temp.seat=local;
temp.di=1;
Push (S,temp);
if (Equal(local, end))
return OK;
local = NextPos (local ,1);
step++;
}
else
{
if (!StackEmpty(*S))
{
Pop(S,&temp);
while (temp.di==4&&!StackEmpty(*S))
{
MarkPrint (maze,temp.seat);
Pop (S,&temp);
step--;
}
if (temp.di<4)
{
temp.di++;
Push (S,temp);
local=NextPos(temp.seat, temp.di);
}
}
}
}while (!StackEmpty (*S));
return ERROR;
}
int Equal (PosType A,PosType B)
{
if (A.x==B.x&&A.y==B.y)
return 1;
else
return 0;
}
int main ()
{
PosType start,end;
SqStack s;
int result=0;
int maze [10][10] =
{ {1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 0, 0, 1, 0, 0, 0, 1, 0, 1},
{1, 0, 0, 1, 0, 0, 0, 1, 0, 1},
{1, 0, 0, 0, 0, 1, 1, 0, 0, 1},
{1, 0, 1, 1, 1, 0, 0, 0, 0, 1},
{1, 0, 0, 0, 1, 0, 0, 0, 0, 1},
{1, 0, 1, 0, 0, 0, 1, 0, 0, 1},
{1, 0, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 0, 0, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}};
start.x=1;
start.y=1;
end.x=8;
end.y=8;
InitStack (&s);
result=MazePath (maze, start, end, &s);
if (result)
{
printf ("found\n");
ShowAll(&s);
}
else
printf ("not found");
getchar();
return 0;
}

阅读(483) | 评论(0) | 转发(0) |

上一篇：栈的应用

下一篇：稀疏矩阵的存储与快速转置