矩阵链乘

C - 矩阵链乘Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld& %lluSubmit StatusDescriptionSuppose 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 printone 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

思路：

就是不断模仿进栈与出栈，难点是当时没想到要用结构体来存储遍历时的字母，所以只是存储了下标，之后就卡住了。因为还要将用了的坐标再压栈重用。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;struct S{int ln,rn;S(){ln=0,rn=0;}}s[26],st[5000];int main(){/*freopen("input.txt","r",stdin);*/int n,i;char c[4],str[5000];scanf("%d",&n);for(i=0;i<n;++i){scanf(" %s",&c[0]);scanf("%d%d",&s[c[0]-'A'].ln,&s[c[0]-'A'].rn);}while(~scanf("%s",&str)){int top=-1,flag=1,ans=0;for(i=0;str[i];++i){if(str[i]>='A'&&str[i]<='Z')st[++top]=s[str[i]-'A'];else if(str[i]==')'){if(st[top].ln!=st[top-1].rn){flag=0;break;}ans+=st[top-1].ln*st[top-1].rn*st[top].rn;top--;st[top].rn = st[top+1].rn;}}if(!flag)printf("error\n");else printf("%d\n",ans);}return 0;}