POJ2246

Matrix Chain Multiplication

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1756		Accepted: 1134

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

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 } 

Line       = Expression 

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.

题目大意：输出每个示例中，矩阵相乘时所进行的运算次数

解题思路：使用栈对表达式进行处理，对两个矩阵相乘完之后的矩阵进行压栈处理

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


解题代码

#include "stdio.h"
#include <ctype.h>
#include <stack>

char m[30][3];
char n[30][3];
int num=0;
void reset();
int countMartrix(int a,int b);
int findMatrix(char A);
int main(){
#ifdef TEST
    freopen("test.txt","r",stdin);
    freopen("testout.txt","w",stdout);
#endif

    scanf("%d",&num);
    int i=0;
    for (i=0;i<num;i++)
    {
        getchar();
        scanf("%c %d %d",&m[i][0],&m[i][1],&m[i][2]);
    }
    getchar();
    int resu=0;
    i=0;
    char buf[100],tmp1,tmp2;
    std::stack<char> s;
    while (fgets(buf,100,stdin)!=NULL)
    {
        reset();//对原来的各个矩阵的行和列数进行保存
        while(buf[i]!='\0'){
            if(buf[i]==')'){
                tmp1=s.top();
                s.pop();
                tmp2=s.top();
                s.pop();
                int tmp1_pos=0,tmp2_pos=0;
                tmp1_pos=findMatrix(tmp1);//找到矩阵在数组中的位置
                tmp2_pos=findMatrix(tmp2);
                if (tmp1_pos>=0&&tmp2_pos>=0){
                    if(countMartrix(tmp2_pos,tmp1_pos)==-1){
                        resu=-1;
                        break;}
                    else resu+=countMartrix(tmp2_pos,tmp1_pos);
                }
                s.pop();
                n[tmp2_pos][2]=n[tmp1_pos][2];//更新相乘过后的矩阵的列数，然后压栈处理
                s.push(tmp2);
            }
            else s.push(buf[i]);
            i++;
        }
        if(resu>=0)
            printf("%d\n",resu);
        else printf("error\n");
        resu=0;
        i=0;
        while(s.empty()==false)
            s.pop();
    }


    return 0; 
}

int countMartrix(int a,int b){//判断矩阵是否可以相乘
    if (n[a][2]==n[b][1])
        return n[a][1]*n[a][2]*n[b][2];
    else
        return -1;
}
int findMatrix(char a){//找到数组中矩阵的位置
    bool ifind;
    int i=0;
    for (i=0;i<num;i++)
    {
        if (m[i][0]==a){
            ifind=true;
            break;}
    }
    if(ifind)
        return i;
    return -1;
}
void reset(){
    for (int i=0;i<num;i++)
    {
        n[i][0]=m[i][0];
        n[i][1]=m[i][1];
        n[i][2]=m[i][2];
    }
}