uva 442 Matrix Chain Multiplication

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原题：
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

中文：

给你一堆矩阵，然后给你一个矩阵乘法的顺序，让你计算按照这个顺序计算矩阵乘法，里面要计算多少次？

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

map<char,pii> mc;
int n;
string s;

int main()
{

    ios::sync_with_stdio(false);
    while(cin>>n)
    {
        mc.clear();
        char a;
        int r,c;
        for(int i=1;i<=n;i++)
        {
            cin>>a>>r>>c;
            mc[a]={make_pair(r,c)};
        }
        while(cin>>s)
        {
            int ans=0;
            int flag=0;
            stack<char> sc;
            stack<pii> mat;
            pii p1,p2;
            for(int i=0;i<s.size();i++)
            {
                if(s[i]=='(')
                    sc.push('(');
                if(s[i]==')')
                {
                    p1=mat.top();
                    mat.pop();

                    p2=mat.top();
                    mat.pop();
                    if(p1.first!=p2.second)
                    {
                        flag=1;
                        break;
                    }
                    ans+=p1.second*p1.first*p2.first;
                    mat.push(make_pair(p2.first,p1.second));
                    sc.pop();
                }
                if(s[i]!='('&&s[i]!=')')
                    mat.push(mc[s[i]]);
            }
            if(flag)
                cout<<"error"<<endl;
            else
                cout<<ans<<endl;
        }
    }
    return 0;
}

解答：

小白书的基础题目，主要就是如何处理乘法序列的表达式。

利用两个栈，一个用来存储括号，另一个用来存储矩阵，每次遍历表达式，遇到左括号和矩阵时分别入栈，遇到右括号时候，计算当前栈顶的两个矩阵乘法，再将乘法后的矩阵入栈，如此下来即可。