【STL】——乱搞——UVA442

                                                                                              List_4

                                         Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

                       

Description

 

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

题意：计算矩阵相乘的时候需要计算的乘法的次数。

先放上正常版的代码：

//判断能否矩阵运算，不能则报错，可以则计算需要运算多少次乘法，每次遇到右括号弹栈并判断
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<stack>
using namespace std;
struct Matrices
{
    int l;
    int r;
}M[31];
int main()
{
    int T;
    cin>>T;
    memset(M,0,sizeof(M));
    char ch[2];
    while(T--)
    {
        cin>>ch;
        cin>>M[ch[0]-'A'].l>>M[ch[0]-'A'].r;
    }
    string a;
    stack<Matrices> S;
    while(cin>>a)
    {
        int ans=0,flag=1;
        int len=a.length();
        for(int i=0;i<len;i++)
        {
            if(a[i]=='(') continue;
            else if(a[i]==')')
            {
                M[27].l=S.top().l;
                M[27].r=S.top().r;
                S.pop();
                M[28].l=S.top().l;
                M[28].r=S.top().r;
                S.pop();
                if(M[28].r!=M[27].l)
                {
                    flag=0;
                    break;
                }
                ans+=M[28].l*M[28].r*M[27].r;
                M[29].l=M[28].l;
                M[29].r=M[27].r;
                S.push(M[29]);
            }
            else S.push(M[a[i]-'A']);
        }
        if(!flag) printf("error\n");
        else printf("%d\n",ans);
    }
    return 0;
}

 

下面是一只笨到家的妖怪的代码：（TAT）

#include<bits/stdc++.h>
using namespace std;

struct mat{
    //char id;
    int l,r;
}temp,change;

int main()
{
    map<char,pair<int,int> >p;

    int n;
    cin>>n;
    char idnum;
    int r,c;
    for(int i=0;i<n;i++)
    {
        cin>>idnum>>r>>c;
        p[idnum].first=r;
        p[idnum].second=c;
    }

    string op;
    stack<mat>s;

    while(cin>>op)
    {
        int sum=0;
        int i;
        for(i=0;i<op.length();i++)
        {
            if(op[i]=='(')continue;
            else if(isalpha(op[i]))
            {
                temp.l=p[op[i]].first;
                temp.r=p[op[i]].second;
                s.push(temp);
            }
            else if(op[i]==')')
            {
                temp=s.top();
                s.pop();
                if(s.top().r!=temp.l)
                {
                    break;
                }
                sum+=(s.top().l*s.top().r*temp.r);
                temp.l=s.top().l;
                s.pop();
                s.push(temp);
            }
        }
        if(i!=op.length())
        {
            cout<<"error"<<endl;
        }
        else
        {
            cout<<sum<<endl;
        }
    }
    return 0;
}