uva-442 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
0 0 0 error 10000 error 3500 15000 40500 47500 15125
题目意思：给出字符代表矩阵的定义，计算表达示中矩阵相乘的总次数；
解题思路：定义一个stack<node>st,用node（－1，－1）代表左括号,从左到右扫描，如果是左括号则st.push(node(-1,01));如果是字母则push字母对应的矩阵；如果是右括号则把矩阵pop出来计算，一直到左括号；
#include <iostream>
#include<deque>
#include<algorithm>
#include<cstdio>
#include<stack>
#include<string>
#include<vector>
#include<map>
using namespace std;
//bool check (vector<string>&v,map<string,int>&m)
//{
// for(unsigned int i=1;i<v.size();i++)
// if(m[v[i]]-m[v[i-1]]!=1)
// return false;
// return true;
//}
//int main()
//{
// int cas;
// cin>>cas;
// getchar();
// while(cas--)
// {
// int n;
// cin>>n;
// getchar();
// vector<string>v;
// string s;
// map<string ,int>m;
// for(int i=0; i<n; i++)
// {
// getline(cin,s);
// v.push_back(s);
// }
// for(int i=0; i<n; i++)
// {
// getline(cin,s);
// m[s]=i;
// }
// for(int q=0; q<n-1;)
// {
// //if(check(v,m))break;
// bool flag=1;
// for(int i=q+1; i<n; i++)
// {
// if(m[v[q]]-m[v[i]]==1)
// {
// cout<<v[i]<<endl;
// string t=v[i];
// for(int j=i; j>0; j--)
// v[j]=v[j-1];
// v[0]=t;
// flag=0;
// q=0;
// break;
// }
// }
// if(flag)q++;
// }
// cout<<endl;
// }
// return 0;
//}
struct node
{
int x,y;
node(int a=0,int b=0):x(a),y(b) {}
};
int main()
{
int n;
cin>>n;
char c;
int x,y;
node arr[200];
for(int i=0; i<n; i++)
cin>>c>>x>>y,arr[c]=node(x,y);
string exp;
while(cin>>exp)
{
stack<node>st;
int sum=0;
bool flag=0;
for(int i=0; i<exp.size(); i++)
{
if(exp[i]!=')')
{
if(exp[i]=='(')
st.push(node(-1,-1));
else
st.push(arr[exp[i]]);
}
else if(!st.empty())
{
node matrix1=st.top();
st.pop();
while(st.top().x!=-1)
{
node matrix2=st.top();
st.pop();
if(matrix1.x!=matrix2.y)
{
flag=1;
cout<<"error"<<endl;
i=exp.size();
break;
}
sum+=matrix2.x*matrix2.y*matrix1.y;
matrix1=node(matrix2.x,matrix1.y);
}
st.pop();
st.push(matrix1);
}
else
{
cout<<"error"<<endl;
flag=1;
}
}
while(st.size()!=1&&!flag)
{
node matrix2=st.top();
st.pop();
node matrix1=st.top();
st.pop();
if(matrix1.y!=matrix2.x)
{
cout<<"error"<<endl;
break;
}
st.push(node(matrix1.x,matrix2.y));
sum+=matrix1.x*matrix1.y*matrix2.y;
}
if(!flag)
cout<<sum<<endl;
}
return 0;
}