ZOJ-1094-Matrix Chain Multiplication
2017-04-19 20:41
369 查看
Matrix multiplication problem is a typical example of dynamical programming.
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):
Otherwise print one line containing the number of elementary multiplications needed to evaluate the expression in the way specified by the parentheses.
使用栈
#include <iostream>
#include <stack>
#include <string>
using namespace std;
struct Matrix
{
char name;
int row,column;
};
int main()
{
stack <Matrix> m;
Matrix store[26];
int num_of_Matrix;
while(cin>>num_of_Matrix)
{
for(int i=0;i<num_of_Matrix;i++)
cin>>store[i].name>>store[i].row>>store[i].column;
string str;
while(cin>>str)
{
int result=0;
int i=0;
int judge=1;
while(str[i++]!='\0')
{
if(str[i]!='('&&str[i]!=')')
{
for(int j=0;j<num_of_Matrix;j++)
if(store[j].name==str[i])
{
m.push(store[j]);
}
}
else if(str[i]==')')
{
Matrix n,p;
int r;
p=m.top();
n.column=m.top().column;
r=m.top().row;
m.pop();
if(r!=m.top().column)
{
m.push(p);
judge=0;
break;
}
n.row=m.top().row;
m.pop();
m.push(n);
result+=n.column*n.row*r;
}
else
continue;
}
if(!m.empty())
m.pop();
if(judge==0)
cout<<"error"<<endl;
else
cout<<result<<endl;
}
}
return 0;
}
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 (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 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 4750015125
使用栈
#include <iostream>
#include <stack>
#include <string>
using namespace std;
struct Matrix
{
char name;
int row,column;
};
int main()
{
stack <Matrix> m;
Matrix store[26];
int num_of_Matrix;
while(cin>>num_of_Matrix)
{
for(int i=0;i<num_of_Matrix;i++)
cin>>store[i].name>>store[i].row>>store[i].column;
string str;
while(cin>>str)
{
int result=0;
int i=0;
int judge=1;
while(str[i++]!='\0')
{
if(str[i]!='('&&str[i]!=')')
{
for(int j=0;j<num_of_Matrix;j++)
if(store[j].name==str[i])
{
m.push(store[j]);
}
}
else if(str[i]==')')
{
Matrix n,p;
int r;
p=m.top();
n.column=m.top().column;
r=m.top().row;
m.pop();
if(r!=m.top().column)
{
m.push(p);
judge=0;
break;
}
n.row=m.top().row;
m.pop();
m.push(n);
result+=n.column*n.row*r;
}
else
continue;
}
if(!m.empty())
m.pop();
if(judge==0)
cout<<"error"<<endl;
else
cout<<result<<endl;
}
}
return 0;
}
相关文章推荐
- ZOJ1094-Matrix Chain Multiplication
- zoj1094 Matrix Chain Multiplication
- ZOJ 1094 Matrix Chain Multiplication(map函数 模拟)
- POJ2246 HDU1082 ZOJ1094 UVA442 Matrix Chain Multiplication(矩阵相乘)
- UVa 442 - Matrix Chain Multiplication(zoj 1094)
- POJ2246 HDU1082 ZOJ1094 UVA442 Matrix Chain Multiplication题解
- zoj1094 Matrix Chain Multiplication(模拟)
- zoj 1094 Matrix Chain Multiplication
- 【zoj1094-Matrix Chain Multiplication】
- zoj1094 Matrix Chain Multiplication 模拟
- zoj-1094-Matrix Chain Multiplication
- zoj 1094 poj 2246 Matrix Chain Multiplication(堆栈)
- zoj1094-Matrix Chain Multiplication
- ZOJ Problem Set - 1094 Matrix Chain Multiplication
- ZOJ 1094_Matrix Chain Multiplication
- zoj 1094 poj 2246 Matrix Chain Multiplication(堆栈)
- ZOJ-1094,POJ-2246 Matrix Chain Multiplication
- ZOJ 1094 Matrix Chain Multiplication
- zoj-1094Matrix Chain Multiplication(栈来实现矩阵相乘)
- ZOJ-1094-Matrix Chain Multiplication