题目例如以下：

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 Specification

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 } <EOF>
Line       = Expression <CR>
Expression = Matrix | "(" Expression Expression ")"
Matrix     = "A" | "B" | "C" | ... | "X" | "Y" | "Z"

Output Specification

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

我用这道题练了练STL库中的map和pair，感觉熟悉了很多，一遍AC了。我是直接模拟的，遇到括号内有两个字母的情况，直接raplace成一个新矩阵（用小写表示），并给count加上乘法的数目，遇到左行不等于右列的情况，跳出循环，输出error。

AC的代码例如以下：

#include 
#include
#include
#include
#include
using namespace std;
map > matrix;
int main()
{
    int n;
    cin>>n;
    getchar();
    char c;
    int d1,d2;
    while(n--)
    {
        cin>>c>>d1>>d2;
        getchar();
        pairm(d1,d2);

        matrix.insert(make_pair(c,m));


    }
    string s;
    char f='a';
    while(cin>>s)
    {

        string::iterator i,j,j2;
        int cou=0,ok=1,flag=1;

        while(ok==1&&flag==1)
        {
            ok=0;
            for(i=s.begin(); i!=s.end(); i++)
            {
                if(*i=='(')
                {
                    j=i;
                    j++;
                    if(isalpha(*j))
                    {
                        j++;
                        if(isalpha(*j))
                        {
                            ok=1;
                            j=i;
                            j++;
                            map >::iterator iter;
                            iter=matrix.find(*j);
                            j++;
                            map >::iterator iter2;
                            iter2=matrix.find(*j);
                            if((iter->second).second!=(iter2->second).first)
                            {
                                flag=0;
                                break;
                            }
                            cou+=((iter->second).first)*((iter2->second).second)*((iter->second).second);
                            j=i;
                            j2=j++;
                            j++;
                            j++;
                            j++;

                            s.replace(j2,j,1,f);
                            pairm3((iter->second).first,(iter2->second).second);

                            matrix.insert(make_pair(f++,m3));
                            }
                    }
                }


            }
        }
        if(flag==0)
            cout<<"error"<