poj 1573 Robot Motion

题意：已知坐标系大小，和初始的最上面的位置，根据命令求是否能出去，或者一直循环

这就是我建的系，画的有点难看哈！

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<stack>
 6 #include<queue>
 7 #include<cmath>
 8 #include<algorithm>
 9 using namespace std;
10 char map[500][500];
11 int vis[500][500];
12 
13 int dx[5]={0,0,-1,1};//分别代表N,S,W,E
14 int dy[5]={1,-1,0,0};
15 int main()
16 {
17     int i,j,a,b,t,f,sum,x;
18     while(cin>>a>>b>>t&&(a!=0||b!=0||t!=0))
19     {
20         sum=0;
21         memset(vis,0,sizeof(vis));
22         for(j=a; j>=1; j--)
23         {
24             for(i=1; i<=b; i++)
25             cin>>map[i][j];
26         }
27         i = t; j = a;
28         while(1)
29         {
30             sum++;
31             vis[i][j]=sum;
32             if(map[i][j]=='N') {i+=dx[0]; j+=dy[0];}
33             else if(map[i][j]=='S') {i+=dx[1]; j+=dy[1];}
34             else if(map[i][j]=='W') {i+=dx[2]; j+=dy[2];}
35             else if(map[i][j]=='E') {i+=dx[3]; j+=dy[3];}
36              if(vis[i][j])
37             {
38                 f=1;
39                 x=vis[i][j];
40                 break;
41             }
42             if(i>b||j>a || i<=0 ||j<=0)
43             {
44                 f=2;
45                 break;
46             }
47         }
48        if(f==2)
49        {
50            printf("%d step(s) to exit
",sum);
51        }
52        else
53        {
54            printf("%d step",x-1);
55            printf("(s)");
56            printf(" before a loop of %d step",sum-x+1);
57            printf("(s)");
58            printf("
");
59        }
60     }
61     return 0;
62 }