hdu 1082 Matrix Chain Multiplication

题目地址：

http://acm.hdu.edu.cn/showproblem.php?pid=1082

题目描述：

Matrix Chain Multiplication

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 963 Accepted Submission(s): 659



Problem Description

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

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：

计算矩阵相乘运算中一共有多少次乘法。

题解：

栈计算表达式。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;
typedef struct Matrix
{
int row;
int col;
char c;
Matrix()
{
row=0;
col=0;
c='\0';
}
}Matrix,*MatrixLink;
Matrix Mats[26];//the matrix lib
Matrix stack_ex[1000];//the expression stack
int top=-1;
int n=0;//the number of the matrix
char query_str[1000]={'\0'};
/*for test*/
int test()
{
return(0);
}
/*calculate the matrix*/
bool CalcMa(Matrix &m1,Matrix &m2,Matrix &m3)
{
if(m1.col!=m2.row)
{
return(false);
}
else
{
m3.row=m1.row;
m3.col=m2.col;
return(true);
}
}
/*calculate the matrix multiplications */
int CalcQue(char que[])
{
int num=0;
top=-1;
int i=0;
while(que[i]!='\0')
{
Matrix NewMa;
switch(que[i])
{
case '('://push the stack
top++;
stack_ex[top].c='(';
break;
case ')'://pop the stack until the first '('(include it),and calculate the AB matrix multiplication,and tne new matrix updated.
if(CalcMa(stack_ex[top-1],stack_ex[top],NewMa))
{
num+=stack_ex[top-1].row*stack_ex[top-1].col*stack_ex[top].col;
top-=2;
stack_ex[top].row=NewMa.row;
stack_ex[top].col=NewMa.col;
//printf("%s %d\n",que,num);
}
else
{
num=-1;
return(num);
}
break;
default://push the stack
top++;
stack_ex[top].c=que[i];
stack_ex[top].row=Mats[que[i]-65].row;
stack_ex[top].col=Mats[que[i]-65].col;
break;
}
i++;
}
return(num);
}
/*main process*/
int MainProc()
{
//initialize
scanf("%d",&n);
int i=0,row=0,col=0;
char c='\0';//the name of the matrix
for(i=0;i<=n-1;i++)
{
scanf("%s%d%d",query_str,&row,&col);
c=query_str[0];
//save to the matrix lib
Mats[c-65].row=row;
Mats[c-65].col=col;
Mats[c-65].c=c;
}
//deal with the query
while(scanf("%s",query_str)!=EOF)
{
if(query_str[0]=='\0')
{
continue;
}
int ele_num=CalcQue(query_str);
if(ele_num>=0)
{
printf("%d\n",ele_num);
}
else
{
printf("error\n");
}
}
return(0);
}
int main()
{
MainProc();
return(0);
}