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C语言数据结构之迷宫问题

作者:lear_scu

这篇文章主要为大家详细介绍了C语言数据结构之迷宫问题,具有一定的参考价值,感兴趣的小伙伴们可以参考一下

本文实例为大家分享了数据结构c语言版迷宫问题栈实现的具体代码,供大家参考,具体内容如下

程序主要参考自严蔚敏老师的数据结构c语言版,在书中程序的大体框架下进行了完善。关于迷宫问题的思路可查阅原书。

#include<iostream>
 
using namespace std;
 
#define MAXSIZE 10
typedef int Status;
typedef struct{
 int x;
  int y;
}Postype;
typedef struct{
 int ord;
 Postype seat;
 int dir;
}SElemType;//栈的元素类型
typedef struct{
 //SElemType data[MAXSIZE];
 SElemType* top;
 SElemType* base;
}Stack;//栈的结构类型
typedef struct{
 char arr[MAXSIZE][MAXSIZE];
}MAZETYPE;//迷宫结构体
 
MAZETYPE maze;
void InitMaze()
{
 maze.arr[0][0] = maze.arr[0][1] = maze.arr[0][2] = maze.arr[0][3] = maze.arr[0][4] = maze.arr[0][5] = maze.arr[0][6] = maze.arr[0][7] = maze.arr[0][8] = maze.arr[0][9] = '1';
 maze.arr[1][0] = maze.arr[1][3] = maze.arr[1][7] = maze.arr[1][9] = '1';
 maze.arr[1][1] = maze.arr[1][2] = maze.arr[1][4] = maze.arr[1][5] = maze.arr[1][6] = maze.arr[1][8] = '0';
 maze.arr[2][0] = maze.arr[2][3] = maze.arr[2][7] = maze.arr[2][9] = '1';
 maze.arr[2][1] = maze.arr[2][2] = maze.arr[2][4] = maze.arr[2][5] = maze.arr[2][6] = maze.arr[2][8] = '0';
 maze.arr[3][0] = maze.arr[3][5] = maze.arr[3][6] = maze.arr[3][9] = '1';
 maze.arr[3][1] = maze.arr[3][2] = maze.arr[3][3] = maze.arr[3][4] = maze.arr[3][7] = maze.arr[3][8] = '0';
 maze.arr[4][0] = maze.arr[4][2] = maze.arr[4][3] = maze.arr[4][4] = maze.arr[4][9] = '1';
 maze.arr[4][1] = maze.arr[4][5] = maze.arr[4][6] = maze.arr[4][7] = maze.arr[4][8] = '0';
 maze.arr[5][0] = maze.arr[5][4] = maze.arr[5][9] = '1';
 maze.arr[5][1] = maze.arr[5][2] = maze.arr[5][3] = maze.arr[5][5] = maze.arr[5][6] = maze.arr[5][7] = maze.arr[5][8] = '0';
 maze.arr[6][0] = maze.arr[6][2] = maze.arr[6][6] = maze.arr[6][9] = '1';
 maze.arr[6][1] = maze.arr[6][3] = maze.arr[6][4] = maze.arr[6][5] = maze.arr[6][7] = maze.arr[6][8] = '0';
 maze.arr[7][0] = maze.arr[7][2] = maze.arr[7][3] = maze.arr[7][4] = maze.arr[7][6] = maze.arr[7][9] = '1';
 maze.arr[7][1] = maze.arr[7][5] = maze.arr[7][7] = maze.arr[7][8] = '0';
 maze.arr[8][0] = maze.arr[8][1] = maze.arr[8][9] = '0';
 maze.arr[8][2] = maze.arr[8][3] = maze.arr[8][4] = maze.arr[8][5] = maze.arr[8][6] = maze.arr[8][7] = maze.arr[8][8] = '0';
 maze.arr[9][0] = maze.arr[9][1] = maze.arr[9][2] = maze.arr[9][3] = maze.arr[9][4] = maze.arr[9][5] = maze.arr[9][6] = maze.arr[9][7] = maze.arr[9][8] = maze.arr[9][9] = '1';
}
Status initStack(Stack &s)
{
 s.base = (SElemType*)malloc(MAXSIZE*sizeof(SElemType));
 if (!s.base) return 0;
 s.top = s.base;
 return 1;
}
void Push(Stack &s, SElemType e)
{
 *s.top++ = e;
}
void Pop(Stack &s, SElemType &e)
{
 e = *--s.top;
}
Status StackEmpty(Stack &s)
{
 if (s.top == s.base) return 1;
 else return 0;
}
Status Pass(Postype curpos)
{
 if (maze.arr[curpos.x][curpos.y] == '0')
 return 1;
 else return 0;
}
void Foot(Postype curpos)
{
 maze.arr[curpos.x][curpos.y] = '*';
}
void MarkPrint(Postype curpos)
{
 maze.arr[curpos.x][curpos.y] = '!';
}
Status StructCmp(Postype a, Postype b)
{
 if (a.x = b.x&&a.y == b.y) return 1;
 else return 0;
}
//下一个位置
Postype NextPos(Postype CurPos, int Dir)
{
 Postype ReturnPos;
 switch (Dir)
 {
 case 1:
 ReturnPos.x = CurPos.x;
 ReturnPos.y = CurPos.y + 1;
 break;
 case 2:
 ReturnPos.x = CurPos.x + 1;
 ReturnPos.y = CurPos.y;
 break;
 case 3:
 ReturnPos.x = CurPos.x;
 ReturnPos.y = CurPos.y - 1;
 break;
 case 4:
 ReturnPos.x = CurPos.x - 1;
 ReturnPos.y = CurPos.y;
 break;
 }
 return ReturnPos;
}
 
Status MazePath(Postype start, Postype end)
{
 Stack s;
 SElemType e;
 initStack(s);
 Postype curpos = start;
 int curstep = 1;
 do{
 if (Pass(curpos))
 {
  Foot(curpos);
  e = { curstep, curpos, 1 };
  Push(s, e);
  if (StructCmp(curpos, end)) return 1;
  curpos = NextPos(curpos, 1);
  curstep++;
 }
 else
 {
  if (!StackEmpty(s))
  {
  Pop(s, e);
  while (e.dir ==4 &&!StackEmpty(s))
  {
   MarkPrint(e.seat); Pop(s, e);
  }
  if (e.dir < 4 && !StackEmpty(s))
  {
   e.dir++;
   Push(s, e);
   curpos = NextPos(e.seat, e.dir);
  }
  }
 
 
 }
 
 } while (!StackEmpty(s));
 return 0;
 
}
 
int main()
{
 InitMaze();
 Postype s, e;
 s.x = s.y = 1;
 e.x = e.y = 8;
 if (MazePath(s, e))
 printf("迷宫成功解密!\n");
 else
 printf("解密失败\n");
 for (int i = 0; i < 10; i++)
 {
 for (int j = 0; j < 10; j++)
  {
  printf("%c ", maze.arr[i][j]);
  }
 printf("\n");
  }
 cout << "-=================================" << endl;
 for (int i = 0; i < 10; i++)
 {
 for (int j = 0; j < 10; j++)
 {
  if (maze.arr[i][j] == '*' || maze.arr[i][j] == '!')
  printf("%c ", maze.arr[i][j]);
  else cout << " ";
 }
 printf("\n");
 }
}

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持脚本之家。

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