题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1082
Matrix Chain Multiplication
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1382 Accepted Submission(s): 905
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.
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"
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))
0
0
error
10000
error
3500
15000
40500
47500
15125
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<iostream>
using namespace std;
#define N 30
#define M 50000
struct Mnode{
int m;
int n;
}mm[N];
Mnode stk[M]; int main()
{
int t;
while(~scanf("%d",&t))
{
map<char,int> mp;
for(int i = ; i < t; i++)
{
char ch;
int m, n;
getchar();
scanf("%c %d %d",&ch,&m,&n);
mp[ch] = i;
mm[i].m = m;
mm[i].n = n;
}
char ml[];
while(~scanf("%s",ml))
{
int top = ;
int ans = ;
bool flag = ;
for(int i = ; i < strlen(ml); i++)
{
if(ml[i]<='Z'&&ml[i]>='A'){
Mnode tm;
tm.m = mm[mp[ml[i]]].m;
tm.n = mm[mp[ml[i]]].n;
stk[top++] = tm;
}
else if(ml[i]==')'){
Mnode tm1,tm2,tm3;
tm2 = stk[--top];
tm1 = stk[--top];
if(tm1.n!=tm2.m){ puts("error"); flag = ; break; }
tm3.m = tm1.m;
tm3.n = tm2.n;
ans+=tm1.m*tm1.n*tm2.n;
stk[top++] = tm3;
}
}
if(flag) printf("%d\n",ans);
}
}
return ;
}
刘汝佳大神的代码，注意其中的构造函数的写法，表示如果没有参数的时候自动默认两个参数值都是0
#include<cstdio>
#include<stack>
#include<iostream>
#include<string>
using namespace std;
struct Matrix{
int a, b;
Matrix(int a = , int b = ):a(a),b(b){}
}m[];
stack<Matrix> s; int main()
{
int n;
cin>>n;
for(int i = ; i < n; i++){
string name;
cin>>name;
int k = name[]-'A';
cin>>m[k].a>>m[k].b;
}
string expr;
while(cin>>expr){
int len = expr.length();
bool error = false;
int ans = ;
for(int i = ; i < len; i++){
if(isalpha(expr[i])) s.push(m[expr[i]-'A']);
else if(expr[i]==')'){
Matrix m2 = s.top();s.pop();
Matrix m1 = s.top();s.pop();
if(m1.b!=m2.a){error = true; break;}
ans += m1.a*m1.b*m2.b;
s.push(Matrix(m1.a,m2.b));
}
}
if(error) printf("error\n"); else printf("%d\n",ans);
}
return ;
}