Matrix Chain Multiplication

Matrix Chain Multiplication
Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices.Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary.However, the number of elementary multiplications needed strongly depends on the evaluation order you choose.
For example, let A be a 50*10 matrix, B a 10*20 matrix and C a 20*5 matrix. There are two different strategies to compute A*B*C, namely (A*B)C and A(B*C).The first one takes 15000 elementary multiplications, but the second one only 3500.Your job is to write a program that determines the number of elementary multiplications needed for a given evaluation strategy
Input
Input consists of two parts: a list of matrices and a list of expressions.The first line of the input file contains one integer n (1 ≤ n ≤ 26), representing the number of matrices in the first part. The next n lines each contain one capital letter, specifying the name of the matrix, and two integers, specifying the number of rows and columns of the matrix.The second part of the input file strictly adheres to the following syntax (given in EBNF):
SecondPart = Line { Line }
Line = Expression
Expression = Matrix | “(” Expression Expression “)”
Matrix = “A” | “B” | “C” | … | “X” | “Y” | “Z”
Output
For each expression found in the second part of the input file, print one line containing the word ‘error’if evaluation of the expression leads to an error due to non-matching matrices. Otherwise print one line containing the number of elementary multiplications needed to evaluate the expression in the way specified by the parentheses.

Sample Input
9
A 50 10
B 10 20
C 20 5
D 30 35
E 35 15
F 15 5
G 5 10
H 10 20
I 20 25
A
B
C
(AA)
(AB)
(AC)
(A(BC))
((AB)C)
(((((DE)F)G)H)I)
(D(E(F(G(HI)))))
((D(EF))((GH)I))

Sample Output
0
0
0
error
10000
error
3500
15000
40500
47500
15125

思路：就是求矩阵相乘的乘法次数。用栈来解决（）优先级的问题，其实每个括号里只有两个矩阵，所以遇到’)’就计算栈顶两个矩阵，并将产生的新矩阵压栈。
AC代码：

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<map>#include<stack>using namespace std;const int maxn=10100;map<char,pair<int,int> > m;stack<pair<int,int> >st;int main(){    int n;    while(cin>>n)    {        for(int i=0;i<n;i++)        {            char x;            pair<int,int> p;            cin>>x;            scanf("%d%d",&p.first,&p.second);            m[x]=p;        }       char s[1100];       while(cin>>s)       {           int ans=0,len=strlen(s);           int ok=1;           if(len==1)           {              printf("0\n");//当只有单个矩阵；              continue;           }            for(int i=0;i<len;i++)            {                if(s[i]=='(')                    continue;                if(s[i]>='A'&&s[i]<='Z')                {                    pair<int,int> p=m[s[i]];                    st.push( p );                }                if(s[i]==')'&&st.size()>=2)                    {                        pair<int,int> a,b,c;                        a=st.top();st.pop();                        if(st.empty())//矩阵运算结束；                            break;                        b=st.top();st.pop();                        if(b.second!=a.first)            //注意a是后面那个矩阵，b才是前面的矩阵，（栈的后进先出）                        {                            printf("error\n");                            ok=0;                            break;                        }                        ans+=b.first*b.second*a.second;                        c.first=b.first;                        c.second=a.second;                        //同样因为栈的特性；                        st.push(c);                    }            }             if(ok)               printf("%d\n",ans);       }    }    return 0;}