UVa 442-矩阵链乘 Matrix Chain Multiplication

题目描述

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 ( tex2html_wrap_inline28 ), 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

0
0
0
error
10000
error
3500
15000
40500
47500
15125--------------------------------------------------
分析

1.定义结构体数组,存储矩阵的行列。如,
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

2.定义栈,存放要计算的输入的数据。(括号无需入栈)如,
(AA)
(AB)
(AC)
(A(BC))
((AB)C)
(((((DE)F)G)H)I)
(D(E(F(G(HI)))))
((D(EF))((GH)I))

3.逐个分析要计算的输入字符串,遇到‘)’,出栈并计算把结果存入栈。
代码
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<stack>
 4 #include<string>
 5 using namespace std;
 6 struct mat{
 7         int a,b;
 8         mat(int a=0,int b=0):a(a),b(b){}
 9     } m[26];
10     stack<mat>s;
11 int main()
12 {
13     int n;
14     cin>>n;
15     for(int i=0;i<n;i++)
16     {
17         char q;
18         cin>>q;
19         int k=q-'A';
20         cin>>m[k].a>>m[k].b;
21     }
22     string z;
23     while(cin>>z)
24     {
25         int len=z.length();
26         bool error=0;
27         int ans=0;
28         for(int i=0;i<len;i++)
29         {
30             if(isalpha(z[i])) s.push(m[z[i]-'A']);
31             else if(z[i]==')')
32             {
33                 mat m2=s.top();s.pop();
34                 mat m1=s.top();s.pop();
35                 if(m1.b!=m2.a){
36                     error=1; break;
37                 }
38                 ans+=m1.a*m1.b*m2.b;
39                 s.push(mat(m1.a,m2.b));
40             }    
41         }
42         if(error) cout<<"error"<<endl;
43         else cout<<ans<<endl;
44     }
45     return 0;
46  } 
原文地址:https://www.cnblogs.com/masking-timeflows/p/6767268.html