Matrix Chain Multiplication

Suppose you have to eva luate 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 eva luation 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 eva luation 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 ( ), 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 } 
  
   
Line       = Expression 
   
     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 eva luation of the expression leads to an error due to non-matching matrices. Otherwise print one line containing the number of elementary multiplications needed to eva luate 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

这题的解题思路类似于加减乘除运算

#include 
  
   
#include 
   
     #include 
    
      #include 
      using namespace std; struct mat{ int row; int column; }; stack
      
        st1; stack
       
         st2; map
        
          rows, columns; int main(void){ string s; char c; int r, l; int num; #ifndef ONLINE_JUDGE freopen("f:\\infile.txt","r",stdin); #endif cin >> num; while(num--){ cin >> c >> r >> l; rows[c] = r; columns[c] = l; } cin.ignore(10, '\n'); while(getline(cin, s)){ int flag = 0; int mulTime = 0; while(st2.size()) st2.pop(); for(int i = 0; i < s.size(); i++){ if(s[i] == '('){ st1.push(s[i]); } else if(s[i] == ')'){ mat x, y; y = st2.top(); st2.pop(); x = st2.top(); st2.pop(); if(x.column != y.row){ cout << "error" << endl; flag = 1; break; } mat temp; temp.row = x.row; temp.column = y.column; st2.push(temp); mulTime += (x.row*x.column*y.column); } else{ mat temp; temp.row = rows[s[i]]; temp.column = columns[s[i]]; st2.push(temp); } } if(flag == 0) cout << mulTime << endl; } return 0; }


【大中小】【打印】【繁体】【投稿】【收藏】【推荐】【举报】【评论】【关闭】【返回顶部】
分享到:
上一篇：最短路 Bellman-Ford（贝尔曼-福..	下一篇：表驱动与工厂模式

帐　　号:

密码: (新用户注册)

验证码:

表　　情:

内　　容: