矩阵链乘.cpp

Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Sincematrix multiplicationisassociative, theorderin whichmultiplicationsareperformedis 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 Foreachexpressionfoundinthesecondpartoftheinputfile,printonelinecontainingtheword‘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

#include<iostream>
#include<stack>
#include<queue>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#include<cstdio>
using namespace std;
struct student
{
	int l,r;
	student(){}
	student (int ll,int rr):l(ll),r(rr){}	
}d[1000];
char a[10000];stack<student>A;
int main()
{
	int n;char p;
	scanf("%d",&n);
	for(int i = 0;i < n ;i++)
	{
		getchar();
		cin>>p;
		cin>>d[i].l;
		cin>>d[i].r;
	}
	while(~scanf("%s",&a))
	{
		bool flag = 0;
		int d1 = strlen(a);long long sum=0;
		for(int i = 0;i < d1;i++)
		{
			if('A' <= a[i] && a[i] <= 'Z')
			{
				A.push(d[a[i]-'A']); 
			} 
			else if(a[i] == ')'){
				student t2 = A.top();A.pop();
				student t1 = A.top();A.pop();
				if( t1.r !=  t2.l)
				{
					flag = 1;break;
				}
				else{
					sum += t1.l*t1.r*t2.r;
					A.push(student(t1.l,t2.r));
				}	
			}
		}
		if(flag == 1)
		{
			printf("error\n");
		}
		else{
			printf("%lld\n",sum);
		}
		while(!A.empty())
		{
			A.pop();
		}
	}
	
	return 0;
}