tyvj P1003 越野跑

tyvj P1003 越野跑

如果有足够时间返回，就继续向前走。
以下是我的代码：

#include < iostream >
#include < string >
#define  maxn 100007
using   namespace  std;

int  main()
{
     long  n,m,u,f,d,cnt,ans,r[maxn];
    
    cin >> m >> n >> u >> f >> d;
     for ( long  i = 1 ;i <= n;i ++ )
    {
         string  s;
        cin >> s;
         switch (s.at( 0 ))
        {
             case   ' u ' :r[i] = 1 ; break ;
             case   ' f ' :r[i] = 2 ; break ;
             case   ' d ' :r[i] = 3 ;
        }
    }
     //   Input
    
    cnt = ans = 0 ;
     for ( long  i = 1 ;i <= n;i ++ )
         if (r[i] == 1 || r[i] == 3 )
        {
             if (cnt + u + d <= m)
            {
                cnt += u + d;ans ++ ;
            }
             else   break ;
        }
         else   if (r[i] == 2 )
        {
             if (cnt + f + f <= m)
            {
                cnt += f + f;ans ++ ;
            }
             else   break ;
        }
    
    cout << ans << endl;
     //   Output
return   0 ;
}