Uva 442 Matrix Chain Multiplication (矩阵连乘)

最新推荐文章于 2020-07-14 09:36:53 发布

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本文探讨了矩阵乘法的优化策略，通过合理选择计算顺序，显著减少元素乘法的数量，实现高效矩阵运算。

Matrix Chain Multiplication

Time Limit:Unknown

Memory Limit:Unknown

64bit IO Format:%lld & %llu

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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

题目大意：给出字符代表矩阵的定义，计算表达示中矩阵相乘的总次数

题目解析：如果遇到字母则进栈，如果遇到 ')' 则弹出栈顶两个字符。进行运算：

sum += mat[top2].row * mat[top1].row * mat[top1].col;

再把新的矩阵压入栈中：新矩阵的行和列为

mat[fuck].row = mat[top2].row;
mat[fuck].col = mat[top1].col;

#include <stdio.h>
#include <stack>
#include <string.h>
#include <ctype.h>
using namespace std;

const int N = 1000;
char fuck = '#';
struct Mat {
	int row,col;
};

Mat mat[300];
bool isCaptial(char ch) {
	if(ch >= 'A' && ch <= 'Z')
		return true;
	else
		return false;
}
bool match(char mat1,char mat2) {
	if(mat[mat1].col == mat[mat2].row)
		return true;
	else
		return false;
}
int main() {
	int n;
	scanf("%d",&n); getchar();
	char ch;
	char str[N];
	for(int i = 1; i <= n; i++) {
		scanf("%c",&ch);
		scanf("%d%d",&mat[ch].row,&mat[ch].col);
		getchar();
	}
	mat[fuck].row = 0 , mat[fuck].col = 0;
	stack<char> st;
	while(gets(str)) {
		char top1,top2;
		int i;
		int sum = 0;
		for(i=0; str[i] != '\0'; i++) {
			if(isCaptial(str[i])) {
				st.push(str[i]);
			}else if( str[i] == ')') {
				top1 = st.top();
				st.pop();
				top2 = st.top();
				st.pop();
				if(match(top2,top1)) {
					sum += mat[top2].row * mat[top1].row * mat[top1].col;
					mat[fuck].row = mat[top2].row;
					mat[fuck].col = mat[top1].col;
					st.push(fuck);
					fuck++;
				} else {
					break;
				}
			}
		}
		if(str[i] != '\0')
			printf("error\n");
		else
			printf("%d\n",sum);
	}
	return 0;
}