UVA-Matrix Chain Multiplication（栈）

 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

0
0
0
error
10000
error
3500
15000
40500
47500
15125

题解：矩阵链乘，让求计算矩阵连成后运算的次序；注意There are two different strategies to compute A*B*C, namely (A*B)*C and A*(B*C).也就是说两个相乘必定会出现括号的所以遇见括号不用记录（位置就可以了；

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ");
typedef long long LL;
struct Node{
	int r,l;
	Node(int x=0,int y=0):r(x),l(y){}
};
Node dt[30];
int main(){
	int n,x,y;
	char s[1010];
	scanf("%d",&n);
	mem(dt,0);
	while(n--){
		scanf("%s",s);
		scanf("%d%d",&x,&y);
		dt[s[0]-'A'].r=x;
		dt[s[0]-'A'].l=y;
	}
	while(~scanf("%s",s)){
		stack<Node>S;
		int len=strlen(s);
		Node a,b;
		int ans=0,flot=1;
		for(int i=0;i<len;i++){
			if(isalpha(s[i])){
				S.push(dt[s[i]-'A']);
			}
			else if(s[i]==')'){
				b=S.top();S.pop();
				a=S.top();S.pop();
				//printf("%d %d
",a.l,b.r);
				if(a.l!=b.r){
					flot=0;break;
				}
				ans+=a.r*a.l*b.l;
				S.push(Node(a.r,b.l));
			}
		}
		if(flot)
		printf("%d
",ans);
		else puts("error");
	}
	return 0;
}