zoj 2989 Encoding

题意：<space> = 0, A = 1, B = 2, C = 3, ..., Y = 25, Z = 26 （10进制额度）

     给你字符窜，让你输出数字，每个数字代表二进制的五位数，按一定蛇形顺序排列，然后

     从首位按顺序输出 比如ACM 输入行列都是4 4

最后输出是000010001101101

#include<stdio.h>
#include<stdlib.h>
#include<string.h>   
int vis[500][500];
int a[500][500];
int main()
{
	char Gra[30][6]={"00000","00001","00010","00011","00100","00101","00110","00111",
		           "01000","01001","01010","01011","01100","01101","01110","01111",
				   "10000","10001","10010","10011","10100","10101","10110","10111",
				   "11000","11001","11010"}; 
   int T,c,r;
   char ch1[200];//存储输入的字母
   int ch2[500];
   int ttt=1;
   int i,j;
   scanf("%d",&T);
   while(T--)
   {
	   memset(vis,0,sizeof(vis));
	   scanf("%d%d",&r,&c);
	   getchar();
	   gets(ch1);
	   int len=strlen(ch1);
	   int cas=0;
	   //将acm这样的值变化成数字赋予ch2
	   for (i=0;i<len;i++)//ch1的长度
		   for (j=0;j<5;j++)
		   {
			   if(ch1[i]==' ')  ch2[cas++]=Gra[0][j]-'0';
			   else  ch2[cas++]=Gra[ch1[i]-'A'+1][j]-'0';
		   }
		   //将未填满的位置填充0
		   for (int k=len*5;k<r*c;k++)
			   ch2[cas++]=0;
		   int t=1;
		   int x=0,y=0;//控制位置
		   len=r*c;
           a[0][0]=ch2[0];
		   vis[0][0]=1;
		   while(t<len)
		   {
		      while( y+1<c && !vis[x][y+1]) 
			  {
				  vis[x][++y]=1;
				  a[x][y]=ch2[t++];//右
			  }
			  while( x+1<r && !vis[x+1][y])
			  {
				  vis[++x][y]=1;
				  a[x][y]=ch2[t++];//下
			  }
			  
			  while( y-1>=0 && !vis[x][y-1]) 
			  {
			      vis[x][--y]=1;
				  a[x][y]=ch2[t++];//左
			  }
			  while( x-1>=0 && !vis[x-1][y]) 
			  {
		 	      vis[--x][y]=1;
				  a[x][y]=ch2[t++];//上
			  }
		   }
		   printf("%d ",ttt++);
		   for (i=0;i<r;i++)
			   for (j=0;j<c;j++)
				   printf("%d",a[i][j]);
          printf("\n");
   }
   return 0;
}