ZOJ 1094 Matrix Chain Multiplication

本文探讨了矩阵链乘法问题的动态规划解决方案,详细解释了如何通过优化矩阵的乘法顺序来减少基本乘法运算的数量。通过实例分析了不同乘法顺序导致的基本乘法运算数量差异,并提供了一个使用栈的数据结构实现的算法来计算特定乘法顺序所需的基本乘法运算数。

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Matrix Chain Multiplication

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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 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 (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 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

用stack

#include <iostream>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#define MAX 27
using namespace std;

class matrix_feature{
public:
	int row;
	int col;
};

map<char, matrix_feature> matrix;

int main(){
	int n;
	matrix_feature matrix_prop;
	char name;
	string line;

	cin>>n;
	for(int i = 0; i < n; i++){
		cin>>name>>matrix_prop.row>>matrix_prop.col;
		matrix.insert(make_pair(name, matrix_prop));
		// matrix[name] = matrix_prop;
	}
	// for(map<char,int*>::iterator iter = matrix.begin(); iter != matrix.end(); iter++)
	// 	cout<<iter->first<<' '<<iter->second[0]<<' '<<iter->second[1]<<endl;
	while(cin>>line){
		stack<char> operands;
		stack<matrix_feature> matrix_props;
		// char matrix_now;
		int count = 0, flag = 1;
		matrix_feature matrix_left, matrix_right, matrix_temp;
		for(int i = 0; i < line.size(); i++){
			// while(!operands.empty())
			// 	operands.pop();
			if(line[i] == '(')
				continue;
			else if(line[i] == ')'){
				if( operands.size() == 1){
					break;
				}
				operands.pop();
				matrix_right = matrix_props.top();
				matrix_props.pop();
				operands.pop();
				matrix_left = matrix_props.top();
				matrix_props.pop();
				if(matrix_left.col !=  matrix_right.row){
					cout<<"error"<<endl;
					flag = 0;
					break;
				}
				else{
					count += matrix_left.row * matrix_right.row * matrix_right.col;
					operands.push('T');
					matrix_temp.row = matrix_left.row;
					matrix_temp.col = matrix_right.col;
					matrix_props.push(matrix_temp);
				}
			}
			else{
				operands.push(line[i]);
				matrix_props.push(matrix[line[i]]);
			}
		}
		if(flag)
			cout<<count<<endl;
	}
	return 0;
}


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