UVa-442-Matrix Chain Multiplication

题目：UVa 442 Matrix Chain Multiplication

题目分析：
关键是解析表达式，可以用栈来解决：遇到字母时入栈，遇到右括号时出栈并且计算，然后结果入栈。因为输入保证合法，括号无需入栈。

#include <iostream>
#include <cstdio>
#include <stack>
#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 ; //a capital letter specifying the name of the matrix.
        cin >> name;
        int k = name[0]-'A';
        cin >> m[k].a >> m[k].b;
    }
    string expr;//expression

    while(cin >> expr){
        int len = expr.length();
        bool error = false;
        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
");
        else printf("%d
",ans);

    }



    return 0;
}