C++實現(xiàn)推箱子游戲的案例

#include<stdio.h>
#include<stdlib.h>
#include<getch.h>

int main(int argc,const char*argv[])
{
 int cut=0;
 int a[8][8]={
   {0,0,3,3,3,3,0,0}, 
   {0,0,3,5,5,3,0,0}, 
   {0,0,3,0,5,3,0,0}, 
   {0,3,0,0,4,5,3,0}, 
   {3,3,0,4,0,0,3,3}, 
   {3,0,0,3,4,4,0,3}, 
   {3,0,0,2,0,0,0,3}, 
   {3,3,3,3,3,3,3,3}}; 
 int x=6,y=3;
 for(;;)
 {
 int cunt=0;
 system("clear");
 for(int i=0;i<8;i++)
 {
  for(int j=0;j<8;j++)
  {
  if(a[i][j]==9)
  {
  cunt++;
  }
  switch(a[i][j])
  {
  case 0:printf(" ");break;
  case 2:printf("@ ");break;//人
  case 3:printf("# ");break;//墻
  case 4:printf("$ ");break;//箱子
  case 5:printf("o ");break;//目標位置
  case 7:printf("@ ");break;
  case 9:printf("$ ");break;
  }
  }
  printf("\n");
 }
 if(4==cunt)
 {
  printf("成功\n步數(shù)：%d",cut);
  return 0;
 }
 //根據(jù)數(shù)據(jù)的大小來判斷當前的坐標上的情況
 switch(getch())
 {
 case 183:
   if(0!=x&&3==a[x-1][y])
  {
  a[x][y]=2;
  }
  else if(5==a[x-1][y]||0==a[x-1][y])
  {
  a[x][y]-=2;
  a[x-1][y]+=2;
  x--;
  cut++;
  }
  else if((5==a[x-2][y]||0==a[x-2][y])&&(4==a[x-1][y]||9==a[x-1][y]))
  {
  a[x-2][y]+=4;
  a[x-1][y]-=4;
  a[x-1][y]+=2;
  a[x][y]-=2;
  x--;cut++;
  }break;
 case 184:
  if(0!=x&&3==a[x+1][y])
  {
  a[x][y]=2;
  }
  else if((5==a[x+1][y])||(0==a[x+1][y]))
  {
  a[x][y]-=2;
  a[x+1][y]+=2;
  x++;cut++;
  }
  else if((5==a[x+2][y]||0==a[x+2][y])&&(4==a[x+1][y]||9==a[x+1][y]))
  {
  a[x+2][y]+=4;
  a[x+1][y]-=4;
  a[x+1][y]+=2;
  a[x][y]-=2;
  x++;cut++;
  }break;
 case 185:
   if(0!=x&&3==a[x][y+1])
  {
  a[x][y]=2;
  }
  else if(5==a[x][y+1]||0==a[x][y+1])
  {
  a[x][y]-=2;
  a[x][y+1]+=2;
  y++;cut++;
  }
  else if((0==a[x][y+2]||5==a[x][y+2])&&(4==a[x][y+1]||9==a[x][y+1]))
  {
  a[x][y+2]+=4;
  a[x][y+1]-=4;
  a[x][y+1]+=2;
  a[x][y]-=2;
  y++;cut++;
  }break;
 case 186:
   if(0!=x&&3==a[x][y-1])
  {
  a[x][y]=2;
  }
  else if(5==a[x][y-1]||0==a[x][y-1])
  {
  a[x][y]-=2;
  a[x][y-1]+=2;
  y--;cut++;
  }
  else if((0==a[x][y-2]||5==a[x-1][y])&&(4==a[x][y-1]||9==a[x][y-1]))
  {
  a[x][y-2]+=4;
  a[x][y-1]-=4;
  a[x][y-1]+=2;
  a[x][y]-=2;
  y--;cut++;
  }break;
 }
} 
}