Matrix Chain Multiplication[HDU1082]

Matrix Chain Multiplication

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 834    Accepted Submission(s): 570

 

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

A cost of one multiplication is r1*c1*c2.The qualification of two matrices which can multiply is c1==r2.
In this exercise,the main bother is how to deal with parentheses.I think this is tedious,so I use Depth_Priority_Search.

#include<stdio.h>

#include<string.h>

char str[1500],ch;

int row[256],col[256];

class node

{

public:

    int ro,co,ans;

};

int match(int x)

{

    int lnum=1,rnum=0,i;

    for (i=x+1;i<strlen(str);i++)

    {

        if (str[i]=='(') lnum++;

        if (str[i]==')') rnum++;

        if (lnum==rnum) return i;

    }

}

int find(int l,int r)

{

    int i;

    for (i=l;i<=r;i++)

    if (str[i]=='(') return i;

    return -1;

}

node dfs(int l,int r)

{

    node ret;

    if (l+1==r && str[l]=='(' && str[r]==')')

    {

        ret.ans=0;

        return ret;

    }

    if (str[l]=='(')

    {

        int m=match(l);

        if (m==r) return dfs(l+1,r-1);

        node lans=dfs(l+1,m-1),rans=dfs(m+1,r);

        if (lans.ans==-1 || rans.ans==-1 || lans.co!=rans.ro)

        {

            ret.ans=-1;

            return ret;

        }

        ret.ans=lans.ans+rans.ans+lans.ro*lans.co*rans.co;

        ret.ro=lans.ro;ret.co=rans.co;

        return ret;

    }

    int p=find(l,r);

    if (p>=0)

    {

        node lans=dfs(l,p-1),rans=dfs(p,r);

        if (lans.ans==-1 || rans.ans==-1 || lans.co!=rans.ro)

        {

            ret.ans=-1;

            return ret;

        }

        ret.ans=lans.ans+rans.ans+lans.ro*lans.co*rans.co;

        ret.ro=lans.ro;ret.co=rans.co;

        return ret;

    }

    int i,ro=row[str[l]],co=col[str[l]];

    ret.ans=0;

    ret.ro=row[str[l]];

    ret.co=col[str[r]];

    for (i=l+1;i<=r;i++)

    {

        if (co!=row[str[i]])

        {

            ret.ans=-1;

            return ret;

        }

        ret.ans+=ro*co*col[str[i]];

    }

    return ret;

}

int main()

{

    int i,N;

    scanf("%d",&N);

    ch=getchar();

    for (i=1;i<=N;i++)

    {

        scanf("%c",&ch);

        scanf("%d%d",&row[ch],&col[ch]);

        ch=getchar();

    }

    while (scanf("%s",str)!=EOF)

    {

        int l=0,r=strlen(str)-1;

        node ret=dfs(l,r);

        if (ret.ans==-1) printf("error\n");

        else printf("%d\n",ret.ans);

    }

    return 0;

}