UVA 442 Matrix Chain Multiplication(栈）

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本文介绍了一种使用栈来优化矩阵乘法顺序的方法，通过合理的计算顺序减少乘法次数，提高运算效率。

Question：
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 ( 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
题目大意：矩阵相乘，A：50*10 B:10*20 C：20*5，乘法次数A(BC)=10*20*5（BC相乘）+50*10*5=3500，给你一个序列，输出乘法次数，如果不能相乘则输出“error”
解题思路：这是一个标准栈，将字母一个一个推入栈中，当遇到‘）’符号时，取出栈顶两个元素相乘并推入相乘以后的矩阵，以此类推
(https://www.bnuoj.com/v3/contest_show.php?cid=8307#problem/C)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;
struct node
{
    int x,y;
}a[30];
char c[10000];
int main()
{
    int n;
    scanf("%d",&n);
    char ch;
    for(int i=0;i<30;i++)
        a[i].x=a[i].y=0;
    for(int i=0;i<n;i++)
    {
        getchar();
        int tp1,tp2;
        scanf("%c%d%d",&ch,&tp1,&tp2);
        a[ch-'A'].x=tp1;
        a[ch-'A'].y=tp2;
    }
    stack<node> s;
    while(~scanf("%s",c))
    {
        int len=strlen(c),flag=1,ans=0;
        for(int i=0;i<len;i++)
        {
            if(c[i]>='A'&&c[i]<='Z')
                s.push(a[c[i]-'A']);
            else if(c[i]==')')
            {
                node m1,m2;
                m2=s.top();s.pop();   //此处m1,m2不能取反了，注意思考
                m1=s.top();s.pop();
                if(m1.y!=m2.x)
                {
                    flag=0;
                    break;
                }
                ans+=m1.x*m1.y*m2.y;
                s.push(node{m1.x,m2.y});
            }
        }
        if(flag==0)
            printf("error\n");
        else printf("%d\n",ans);
    }
    return 0;
}