442 - Matrix Chain Multiplication

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 }

<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

输入n个矩阵的维度和一些矩阵链乘表达式，输出乘法的次数。如果乘法无法进行，输出error。假定A是m*n矩阵，B是n*p矩阵，那么AB是m*p矩阵，乘法次数为m*n*p。如果A的列数不等于B的行数，则乘法无法进行。

例如，A是50*10的，B是10*20的，C是20*5的，则(A(BC))的乘法次数为10*20*5（BC的乘法次数）+ 50*10*5（(A(BC))的乘法次数）= 3500。

#include <cstdio>
#include <stack>
#include <string>
#include <iostream>
using namespace std;

// 矩阵结构
struct Matrix{
int first;
int second;
};

// 最多26个矩阵
Matrix m[26];
// 矩阵栈
stack<Matrix> st;

// 矩阵链乘
int matrixMul(string exp) {
// 只有1个矩阵
if(exp.length() == 1) {
return 0;
} else {
int ans = 0;
// 简单的表达式解析
for(int i = 0; i < exp.length(); i++) {
// A-Z进栈
if(exp[i] >= 'A' && exp[i] <= 'Z') {
st.push(m[exp[i] - 'A']);
} else if(exp[i] == ')') {
// )，出栈两个矩阵，算出乘积次数并将结果矩阵入栈
Matrix m2 = st.top();
st.pop();
Matrix m1 = st.top();
st.pop();
// 不能做乘法，返回-1
if(m1.second != m2.first) {
return -1;
}
// 结果累加
ans += m1.first * m1.second * m2.second;
Matrix m3;
m3.first = m1.first;
m3.second = m2.second;
// 结果矩阵入栈
st.push(m3);
}
}
return ans;
}

}

int main() {
int n;
scanf("%d", &n);
// 矩阵输入
for(int i = 0; i < n; i++) {
string s;
int first;
int second;
cin >> s >> first >> second;
m[s[0] - 'A'].first = first;
m[s[0] - 'A'].second = second;
}
string exp;
while(cin >> exp) {
int ans = matrixMul(exp);
if(ans < 0) {
printf("error\n");
} else {
printf("%d\n", ans);
}
}

return 0;
}