hdu 1082 Matrix Chain Multiplication

Matrix Chain Multiplication

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 620    Accepted Submission(s): 426

Problem Description

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

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"

 

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

 

Source

University of Ulm Local Contest 1996

 

//栈的应用

//用栈来处理括号的层数 

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <cmath>
#include <stack>
using namespace std;
struct node
{
    int m,n;
  //  bool f;
};
node hash[200];
char s[1000];
int main()
{
    int i,n,sum;
    char c;
    bool b;
    scanf("%d",&n);
    while(n--)
    {
       getchar();
       scanf("%c",&c);
       scanf("%d%d",&hash[c].m,&hash[c].n);
    }
     node temp,temp1;
    while(scanf("%s",s)!=EOF)
    {   sum=0;b=1;
        stack<char> s1;
        stack<node> s2;
        for(i=0;s[i]!='\0';i++)
        {
            if(s[i]=='(')
               s1.push(s[i]);
            else if(s[i]==')')
                 {
                     c=s1.top();
                     s1.pop();
                     s1.pop();
                     while(!s1.empty()&&s1.top()!='(')
                           {
                              temp=s2.top();
                              s2.pop();
                              temp1=s2.top();
                             if(temp1.n!=temp.m)
                             {
                                 b=0;break;
                             }
                              sum+=temp1.m*temp1.n*temp.n;
                              temp1.n=temp.n;
                              s2.pop();
                              s2.push(temp1);
                              s1.pop();
                            }
                       s1.push(c);
                 }
                 else
                 {
                     if(!s1.empty()&&s1.top()!='(')
                     {
                         temp=s2.top();
                         if(temp.n!=hash[s[i]].m)
                         {
                             b=0;break;
                         }
                         sum+=temp.m*temp.n*hash[s[i]].n;
                         temp.n=hash[s[i]].n;
                         s2.pop();
                         s2.push(temp);
                     }
                     else
                     {
                         s2.push(hash[s[i]]);
                         s1.push('#');
                     }
                 }
        }
        if(b)
          printf("%d\n",sum);
        else
         printf("error\n");
    }
    return 0;
}