HDU 1082 Matrix Chain Multiplication(栈和map的使用)

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

000error10000error3500150004050047500
////  main.cpp//  HDU 1028////  Created by 张嘉韬 on 16/7/30.//  Copyright © 2016年 张嘉韬. All rights reserved.//#include <iostream>#include <cstdio>#include <stack>#include <cmath>#include <cstring>#include <map>//#include <string>//#include <string.h>using namespace std;struct matrix{    long long  r;    long long l;}matrixs[40];map <char,int> m;int main(int argc, const char * argv[]) {    //freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin);    int n;    long long ans=0;    string s;    stack<matrix> solve;    cin>>n;    for(int i=1;i<=n;i++)    {        char tempc;        long long tr,tl;        cin>>tempc>>tr>>tl;        matrixs[i].r=tr;        matrixs[i].l=tl;        m[tempc]=i;    }        while(cin>>s)    {        ans=0;        int flag=1;        long long len=s.length();        if(len==1) {cout<<"0"<<endl; continue;}        for(int i=0;i<len;i++)        {            if(s[i]=='(') continue;            else if(s[i]==')')            {                long long r1,r2,l1,l2;                r2=solve.top().r;                l2=solve.top().l;                solve.pop();                r1=solve.top().r;                l1=solve.top().l;                solve.pop();                //cout<<r1<<"*"<<l1<<" "<<r2<<"*"<<l2<<endl;                if(l1!=r2) {cout<<"error"<<endl;flag=0;break;}                ans+=r1*l1*l2;                matrix temp;                temp.r=r1;                temp.l=l2;                solve.push(temp);            }            else            {                solve.push(matrixs[m[s[i]]]);            }        }        if(flag==1) cout<<ans<<endl;    }    return 0;}

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