UVa 442 Matrix Chain Multiplication

UVa 442 Matrix Chain Multiplication

用栈，每次遇到右括号就把从左括号到右括号之间的部分全部计算。要在读取字符串的时候左边加一个左括号，右边加一个右括号（一开始没有这么做依然AC了，数据太弱了）。
以下是我的代码：

#include < iostream >
#include < string >
#include < stack >
#include < cstdio >
using   namespace  std;
const   int  kMaxn( 27 );

struct  Type
{
    Type():a( 0 ),b( 0 ) {}
    Type( int  aa, int  bb):a(aa),b(bb) {}
     int  a,b;
};
bool   operator != ( const  Type  & a, const  Type  & b)
{
     return  (a.a != b.a  ||  a.b != b.b);
}

int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */

     int  n;
    cin >> n;
    Type r[kMaxn];
     for ( int  i = 1 ;i <= n;i ++ )
    {
         char  ch;
         int  x,y;
        cin >> ch >> x >> y;
        r[ch - ' A ' ].a = x;
        r[ch - ' A ' ].b = y;
    }

     string  s;
     while (cin >> s)
    {
        s.insert( 0 , " ( " );
        s.append( " ) " );
        stack < Type >  q;
         int  ans( 0 );
         bool  success( true );
         for ( int  i = 0 ;i < s.size();i ++ )
        {
             if (s[i] == ' ( ' )
                q.push(Type( - 1 , 0 ));
             else   if (s[i] == ' ) ' )
            {
                Type t;
                 if ( ! q.empty()  &&  q.top() != Type( - 1 , 0 ))
                {
                    t.a = q.top().a;
                    t.b = q.top().b;
                }
                 while ( ! q.empty()  &&  q.top() != Type( - 1 , 0 ))
                {
                    q.pop();
                     if ( ! q.empty()  &&  q.top() != Type( - 1 , 0 ))
                    {
                         if (q.top().b != t.a)
                        {
                            success = false ;
                             break ;
                        }
                         else
                        {
                            ans += q.top().a * q.top().b * t.b;
                            t.a = q.top().a;
                        }
                    }
                }
                q.pop();
                q.push(t);
            }
             else
                q.push(Type(r[s[i] - ' A ' ].a,r[s[i] - ' A ' ].b));
        }
         if (success)
            cout << ans << endl;
         else
            cout << " error " << endl;
    }

     return   0 ;
}