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): 797    Accepted Submission(s): 553

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

9A 50 10B 10 20C 20 5D 30 35E 35 15F 15 5G 5 10H 10 20I 20 25ABC(AA)(AB)(AC)(A(BC))((AB)C)(((((DE)F)G)H)I)(D(E(F(G(HI)))))((D(EF))((GH)I))

 

Sample Output

000error10000error350015000405004750015125
#include<iostream>#include<cstdio>#include<stack>#include<string>using namespace std;struct Tnode{int r,l;}node[30];                     // 记录所给矩阵名称对应的行和列  由于名称从'A'到'Z' 故最好将下标处理成字母对应数字 即'A'对1 'B'对2...char s[1005];            int row[1005],line[1005];      // 模拟栈  将对应的矩阵的行列记录其中  int ans,t,n;bool cal(){int i,j,r1,l1,r2,l2;if(t==1) return true;r1=row[t];l1=line[t];r2=row[t-1];l2=line[t-1];if(l1!=r2) return false;else {ans+=r1*l2*l1;t--;row[t]=r1;line[t]=l2;return true;}}int main(){    int i,j,t1,t2,flag;char c;scanf("%d",&n);for(i=1;i<=n;i++){getchar();scanf("%c%d%d",&c,&t1,&t2);node[c-'A'+1].l=t1;node[c-'A'+1].r=t2;}getchar();    while(gets(s)!=NULL)    {ans=0;t=0;flag=0;        for(i=0;s[i]!='\0';i++){if(s[i]=='(') continue;else if(s[i]>='A'&&s[i]<='Z')            // 为字母则进栈{row[++t]=node[s[i]-'A'+1].r;line[t]=node[s[i]-'A'+1].l;}else if(s[i]=')')                       // 为')'则出栈两个 计算  再进栈一个{if(!cal())                      {flag=1;printf("error\n");     break;}}}if(!flag) printf("%d\n",ans);    }    return 0;}