2013
11-27

# Matrix Chain Multiplication

Matrix multiplication problem is a typical example of dynamical programming.

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 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 } <EOF>
Line = Expression <CR>
Expression = Matrix | "(" Expression Expression ")"
Matrix = "A" | "B" | "C" | … | "X" | "Y" | "Z"

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.

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))

0
0
0
error
10000
error
3500
15000
40500
47500
15125

#include"stdio.h"
#include"string.h"
#include"stack"
using namespace std;

struct node
{
int x,y;
}A[201];

int main()
{
int n;
int i;
int f;
char c;
int ans;
char s[1001];
node t,tt;
scanf("%d",&n);
while(n--)
{
getchar();
scanf("%c",&c);
scanf("%d%d",&A[c].x,&A[c].y);
}
getchar();

while(gets(s))
{
f=1;
ans=0;
stack<char>s1;
stack<node>s2;
for(i=0;s[i];i++)
{
if(s[i]=='(')
s1.push(s[i]);
else if(s[i]==')')
{
c=s1.top();
s1.pop();
s1.pop();
while(!s1.empty()&&s1.top()!='(')
{
tt=s2.top();
s2.pop();
t=s2.top();
if(t.y!=tt.x)
{
f=0;
break;
}
ans+=t.x*t.y*tt.y;
t.y=tt.y;
s2.pop();
s2.push(t);
s1.pop();
}
s1.push(s[i]);
}
else
{
if(!s1.empty()&&s1.top()!='(')
{
t=s2.top();
if(t.y!=A[s[i]].x)
{
f=0;
break;
}
ans+=t.x*t.y*A[s[i]].y;
t.y=A[s[i]].y;
s2.pop();
s2.push(t);
}
else
{
s2.push(A[s[i]]);
s1.push('#');
}
}
}
if(f)printf("%d\n",ans);
else printf("error\n");
}
return 0;
}

1. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术，也不懂什么是算法，从而也不要求程序员懂什么算法，做程序从来不考虑性能问题，只要页面能显示出来就是好程序，这是国内的现状，很无奈。

2. Gucci New Fall Arrivals

This is really nice to know. I hope it will be successful in the future. Good job on this and keep up the good work.