uva442（Matrix Chain Multiplication）—线性表

最新推荐文章于 2021-01-27 17:18:44 发布

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最新推荐文章于 2021-01-27 17:18:44 发布

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分类专栏： uva-线性表文章标签： uva 线性表

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本文探讨了矩阵链乘问题，即如何通过调整乘法顺序来减少基本乘法操作次数。通过具体的例子展示了不同的计算顺序会导致显著的性能差异，并提供了一段C语言实现代码，用于计算给定表达式的最小乘法次数。

Matrix Chain Multiplication

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

模拟矩阵链乘

#include<stdio.h>
#include<string.h>
int row[30],column[30];
typedef struct{
	int row1;
	int column1;
	char c;
}stack;
stack myStack[30];
int top=-1;
int main(){
	int n;
	scanf("%d",&n);
	char c;
	while(n--){
		getchar();
		scanf("%c",&c);
		getchar();
		scanf("%d%d",&row[c-'A'],&column[c-'A']);
	}	
	getchar();
	char t;
	int flag=0;
	int sum=0;
	while((t=getchar())!=EOF){
		if(t=='\n'){
			if(flag==2)
				printf("0\n");
			else if(flag==1){
				printf("error\n");
			}
			else
				printf("%d\n",sum);
			flag=0;
			top=-1;
			sum=0;
		}
		else{
			if(top==-1){
				if(t!='(')
					flag=2;
				else{
					top++;
					myStack[top].column1=0;
					myStack[top].row1=0;
					myStack[top].c=t;
				}
			}
			else{
				if(t=='('){
					top++;
					myStack[top].column1=0;
					myStack[top].row1=0;
					myStack[top].c=t;
				}
			    else if(t==')'){
					int te=myStack[top].column1;
					int temp=myStack[top].column1;
					int ter=myStack[top].row1;
					int tec=myStack[top].column1;
					top--;
					while(myStack[top].c!='('){
						if(ter==myStack[top].column1){
							temp=temp*myStack[top].column1;
							ter=myStack[top].row1;
							tec=myStack[top].column1;
						}
						else
							flag=1;
						top--;
					}
					myStack[top].c='[';
					myStack[top].row1=ter;
					myStack[top].column1=te;
					temp=temp*ter;
					sum=sum+temp;
				}
				else{
					top++;
					myStack[top].column1=column[t-'A'];
					myStack[top].row1=row[t-'A'];
					myStack[top].c=t;
				}
			}
		}
	}		
	return 0;
}