hdu1082 Matrix Chain Multiplication

Matrix Chain Multiplication

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1348 Accepted Submission(s): 884

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 }
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.

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

用栈来解决矩阵乘法问题，碰到左括号不处理，矩阵压栈，碰到右括号时退栈两次，计算需要的乘法运算次数并累加。然后将相乘得到的矩阵压栈。

#include<iostream>#include<stdio.h>#include<cmath>#include<algorithm>#include<string>#include<cstring>#include<string.h>#include<map>#include<queue>#include<stack>#include<list>#include<cctype>#include<fstream>#include<sstream>#include<iomanip>#include<set>#include<vector>#include<cstdlib>#include<time.h>using namespace std;#define mem(x) memset(x,0,sizeof(x))const int INF = 0x3fffffff;const int MAXN = 1000005;struct node{    int row;    int col;};int main(){    int n;    cin >> n;    map<char, node> M;    char c;    for (int i = 0; i < n; i++)    {        cin >> c;        cin >> M[c].row >> M[c].col;    }    string exp;    while (cin >> exp)    {        int count = 0;        stack<node> zhan;        int i;        for ( i = 0; i < exp.size(); i++)        {            if (exp[i] == '(')                continue;            else if (exp[i] == ')')            {                node a, b;                b = zhan.top();                zhan.pop();                a = zhan.top();                zhan.pop();                if (a.col != b.row)                    break;                count += a.row*b.row*b.col;                node c = { a.row, b.col };                zhan.push(c);            }            else            {                zhan.push(M[exp[i]]);            }        }        if (i == exp.size())            cout << count << endl;        else            cout << "error" << endl;    }    return 0;}