HDU 1035 Robot Motion

思路：简单的模拟，按照矩阵中路径给出的指令走就是了，输出走的步数或者循环状态的步数

[cpp]
#include<stdio.h>
#include<string.h>
int map[15][15],a[15][15];
int n,m,k,kind,link;
bool flag;
void Init()
{
   memset(map,0,sizeof(map));
   int i;
   for(i=0;i<=n+1;i++)
   {
     map[i][0]=-1;map[i][m+1]=-1;
   }
   for(i=0;i<=m+1;i++)
   {
      map[0][i]=-1;map[n+1][i]=-1;
   }
}
void dfs(int x,int y)
{
   if(map[x][y]==-1)
   {
      flag=true;
      return;
   }
   map[x][y]=++kind;
   if(a[x][y]==1)
   {

      if(map[x][y+1]>0)
      {
         link=map[x][y]-map[x][y+1]+1;
         kind=map[x][y+1]-1;
         return;
      }
      else
          dfs(x,y+1);
   }
   if(a[x][y]==2)
   {
      if(map[x][y-1]>0)
      {
         link=map[x][y]-map[x][y-1]+1;
         kind=map[x][y-1]-1;
         return;
      }
      else
          dfs(x,y-1);
   }
   if(a[x][y]==3)
   {
       if(map[x+1][y]>0)
      {
         link=map[x][y]-map[x+1][y]+1;
         kind=map[x+1][y]-1;
         return;
      }
      else
          dfs(x+1,y);
   }
   if(a[x][y]==4)
   {
     if(map[x-1][y]>0)
      {
         link=map[x][y]-map[x-1][y]+1;
         kind=map[x-1][y]-1;
         return;
      }
      else
          dfs(x-1,y);
   }

}
int main()
{
int i,j;
char ch[15];
while(scanf("%d%d",&n,&m)!=EOF&&(n+m))
{
     scanf("%d",&k);
     Init();
     for(i=1;i<=n;i++)
     {
        scanf("%s",ch);
         for(j=0;j<m;j++)
         {
            if(ch[j]=='E')
                a[i][j+1]=1;
            if(ch[j]=='W')
                a[i][j+1]=2;
            if(ch[j]=='S')
                a[i][j+1]=3;
            if(ch[j]=='N')
                a[i][j+1]=4;
         }
     }
     kind=0;flag=false;
     dfs(1,k);
     if(flag) printf("%d step(s) to exit\n",kind);
     else printf("%d step(s) before a loop of %d step(s)\n",kind,link);

}
return 0;
}

补充：软件开发 , C++ ,