UVa 442 Matrix Chain Multiplication

最新推荐文章于 2022-12-31 19:39:21 发布

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最新推荐文章于 2022-12-31 19:39:21 发布

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分类专栏：数据结构文章标签： UVA

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本文介绍了一种通过优化计算顺序来减少矩阵链乘法所需的元算次数的方法，并提供了一个具体的实现示例。文章讨论了不同计算顺序对于所需计算量的影响，并提出了一种使用栈结构来跟踪计算过程的有效算法。

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Matrix Chain Multiplication

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 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 } <EOF>
Line       = Expression <CR>
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 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

题意是计算矩阵乘法的元算次数。

考察的是关于栈的知识。

如果当前字符不是括号，就读入栈中，当前括号是右括号时，就对栈顶的两个元素，进行操作，判断乘法次数，

并更新栈顶元素。

代码如下：

#include <stdio.h>
#include <string.h>
char name[30];

int main(void)
{
    
    struct 
    {
     int row;
     int col;
    }stack[1000];
    
    int find(char ch);
    
    int i,j,k,sum;
    int n,row[30],col[30];
    char operat[1000];
    scanf("%d",&n); 
    for(i=0;i<n;i++)   
    {
     getchar();
     scanf("%c",&name[i]);
     scanf("%d%d",&row[i],&col[i]);
    }
    while(scanf("%s",operat)!=EOF)
    {
     int j=0,sum=0,len,flag=0;
     len=strlen(operat);
     if(len==1)
      printf("0\n"); 
     else
     {
      for(i=0;i<len;i++)
      {
       if(operat[i]!='('&&operat[i]!=')')//进栈字符
       {
        k=find(operat[i]);//找到该字符对应字符串中下标 
        stack[j].row=row[k];
        stack[j++].col=col[k];
       }
       else if(operat[i]==')')//两个字符出栈
       {
        if(stack[j-1].row==stack[j-2].col)
        {
         sum+=(stack[j-2].row*stack[j-2].col*stack[j-1].col);
         stack[j-2].col=stack[j-1].col;
         j--;
        }
        else 
        {
         printf("error\n"); 
         flag=1;   
         break;
        }
       }
      }
      if(!flag)
       printf("%d\n",sum);       
     }      
    }    
    return 0;
} 

int find(char ch)
{
    int i;
    for(i=0;i<30;i++)
     if(ch==name[i])
      return i;
}