UVA422

Description
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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 ( tex2html_wrap_inline28 ), 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 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
这道题是利用栈来解的一道题，在查到‘）’时计算，紫书上的题。紫书上有详细解答。

#include <iostream>#include <stack>#include <cstdio>#include <string>using namespace std;struct Matrix {    int a, b;    Matrix (int a = 0, int b = 0) : a(a), b(b) {}} m[26];stack<Matrix> s;int main(){    int n;    cin >> n;    for (int i = 0; i < n; i++) {        string name;        cin >> name;        int k = name[0] - 'A';//用k存储矩阵名字，这样就能比较好的用数组来表示矩阵        cin >> m[k].a >> m[k].b;    }    string expr;    while (cin >> expr) {        int len = expr.length();        bool error = false;//不再用1、0存储是否可以，用bool变量存储        int ans = 0;        for (int i = 0; i < len; i++) {            if (isalpha (expr[i])) s.push(m[expr[i] - 'A']);            else if (expr[i] == ')') {                Matrix m2 = s.top(); s.pop();                Matrix m1 = s.top(); s.pop();                if (m1.b != m2.a) {                    error = true;                    break;                }                ans += m1.a * m1.b * m2.b;                s.push(Matrix(m1.a, m2.b));            }        }        if (error) {            printf ("error\n");        } else {            printf ("%d\n", ans);        }    }    return 0;}