基于C語言實現(xiàn)簡單的走迷宮游戲

更新時間：2016年04月18日 16:07:22 作者：LiaoGlenn

這篇文章主要介紹了基于C語言實現(xiàn)簡單的走迷宮游戲，用到雙向隊列，方便在運行完畢后輸出經(jīng)過的點，文中示例代碼介紹的非常詳細，具有一定的參考價值，感興趣的小伙伴們可以參考一下

本文實例講述了C語言實現(xiàn)簡單的走迷宮游戲的方法，代碼完整，便于讀者理解。

學(xué)數(shù)據(jù)結(jié)構(gòu)時用“?！睂懙囊粋€走迷宮程序，實際上用到雙向隊列，方便在運行完畢后輸出經(jīng)過的點。

#include <cstdio>
#include <deque>
#include <windows.h>
using namespace std;
class node
{
public:
 int x,y;
 int lastOpt;
};
deque<node> sta;
int x,y;
int endx,endy;
int mapW,mapH;
int steps;
int xopt[5]= {0,0,1,0,-1};
int yopt[5]= {0,1,0,-1,0};
int map[100][100]=
{
 
};
void init()
{
 x = 1;
 y = 1;
 endx = 1;
 endy = 9;
 mapH = 10;
 mapW = 10;
 for(int i=0; i<=mapH; i++)
  for(int j=0; j<=mapW; j++)
  {
   if(i==0 ||j==0 ||i==mapH||j==mapW)
    map[i][j]=-1;
  }
 
 steps=0;
 map[1][2]=-1;
 map[2][2]=-1;
 map[3][2]=-1;
 map[4][2]=-1;
 
 map[6][2]=-1;
 map[7][2]=-1;
 map[8][2]=-1;
 map[9][2]=-1;
 map[9][3]=-1;
 map[8][3]=-1;
 map[1][4]=-1;
 map[3][4]=-1;
 map[4][4]=-1;
 map[5][4]=-1;
 map[6][4]=-1;
 map[7][4]=-1;
 map[1][6]=-1;
 map[2][6]=-1;
 map[3][6]=-1;
 map[4][6]=-1;
 map[5][6]=-1;
 map[6][6]=-1;
 map[7][6]=-1;
 map[8][6]=-1;
 map[8][7]=-1;
 map[8][8]=-1;
 map[7][8]=-1;
 map[6][8]=-1;
 map[5][8]=-1;
 map[4][8]=-1;
 map[3][8]=-1;
 map[2][8]=-1;
 map[1][8]=-1;
 
 map[endx][endy]=5;
}
 
void dis()
{
 system("cls");
 int ori = map[x][y];
 map[x][y]=1;
 for(int i=0; i<=mapH; ++i)
 {
  for(int j=0; j<=mapW; ++j)
  {
   if(map[i][j]==0)
    printf(" ");
   else if(map[i][j]==-1)
    printf(" #");
   else if(map[i][j]==1)
    printf(" @");
   else if(map[i][j]==2)
    printf(" .");
   else if(map[i][j]==5)
    printf(" !");
  }
  cout<<i<<endl;
 }
 for(int j=0; j<=mapW; ++j)
  cout<<j<<" ";
 printf("\n\n > steps:%d Exit:(%d,%d)\n",steps,endx,endy);
 map[x][y] = ori;
}
int can(int n)
{
 if(map[x+xopt[n]][y+yopt[n]] == 0 || map[x+xopt[n]][y+yopt[n]] == 5)
  return 1;
 return 0;
 
}
void visit(int n)
{
 map[x][y]=2;
 x+=xopt[n];
 y+=yopt[n];
 node tem;
 tem.x = x;
 tem.y = y;
 tem.lastOpt = n;
 sta.push_back(tem);
 steps++;
}
int main()
{
 init();
 node tem;
 while( x != endx || y!=endy)
 {
 
  int cans = 0;
  for(int i=1; i<=4; i++)
  {
   if(can(i))
   {
    cans = 1;
    visit(i);
    break;
   }
  }
  if(!cans)
  {
   if(!sta.empty())
   {
    tem = sta.back();
    map[tem.x][tem.y]=0;
    sta.pop_back();
   }
   else
   {
    map[x][y]=2;
    x+=xopt[tem.lastOpt];
    x+=yopt[tem.lastOpt];
    dis();
    break;
   }
 
  }
 
  dis();
  Sleep(500);
 
 }
 if(x==endx && y == endy)
 cout<<"\n > i am finished....\n";
 else
 cout<<"\n > i am finished...but i can't find the right way\n";
 return 0;
}

效果圖：