POJ-2246

Matrix Chain Multiplication

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1756		Accepted: 1134

Description

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

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 } 
Line       = Expression 
Expression = Matrix | "(" Expression Expression ")"
Matrix     = "A" | "B" | "C" | ... | "X" | "Y" | "Z"複製代碼

Output

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.

題目大意：輸出每一個示例中，矩陣相乘時所進行的運算次數

解題思路：使用棧對錶達式進行處理，對兩個矩陣相乘完以後的矩陣進行壓棧處理

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))
複製代碼

Sample Output

0
0
0
error
10000
error
3500
15000
40500
47500
15125複製代碼

解題代碼

1 #include "stdio.h"
 2 #include <ctype.h>
 3 #include <stack>
 4 
 5 char m[30][3];
 6 char n[30][3];
 7 int num=0;
 8 void reset();
 9 int countMartrix(int a,int b);
10 int findMatrix(char A);
11 int main(){
12 #ifdef TEST
13     freopen("test.txt","r",stdin);
14     freopen("testout.txt","w",stdout);
15 #endif
16 
17     scanf("%d",&num);
18     int i=0;
19     for (i=0;i<num;i++)
20     {
21         getchar();
22         scanf("%c %d %d",&m[i][0],&m[i][1],&m[i][2]);
23     }
24     getchar();
25     int resu=0;
26     i=0;
27     char buf[100],tmp1,tmp2;
28     std::stack<char> s;
29     while (fgets(buf,100,stdin)!=NULL)
30     {
31         reset();//對原來的各個矩陣的行和列數進行保存
32         while(buf[i]!='\0'){
33             if(buf[i]==')'){
34                 tmp1=s.top();
35                 s.pop();
36                 tmp2=s.top();
37                 s.pop();
38                 int tmp1_pos=0,tmp2_pos=0;
39                 tmp1_pos=findMatrix(tmp1);//找到矩陣在數組中的位置
40                 tmp2_pos=findMatrix(tmp2);
41                 if (tmp1_pos>=0&&tmp2_pos>=0){
42                     if(countMartrix(tmp2_pos,tmp1_pos)==-1){
43                         resu=-1;
44                         break;}
45                     else resu+=countMartrix(tmp2_pos,tmp1_pos);
46                 }
47                 s.pop();
48                 n[tmp2_pos][2]=n[tmp1_pos][2];//更新相乘事後的矩陣的列數，而後壓棧處理
49                 s.push(tmp2);
50             }
51             else s.push(buf[i]);
52             i++;
53         }
54         if(resu>=0)
55             printf("%d\n",resu);
56         else printf("error\n");
57         resu=0;
58         i=0;
59         while(s.empty()==false)
60             s.pop();
61     }
62 
63 
64     return 0; 
65 }
66 
67 int countMartrix(int a,int b){//判斷矩陣是否能夠相乘
68     if (n[a][2]==n[b][1])
69         return n[a][1]*n[a][2]*n[b][2];
70     else
71         return -1;
72 }
73 int findMatrix(char a){//找到數組中矩陣的位置
74     bool ifind;
75     int i=0;
76     for (i=0;i<num;i++)
77     {
78         if (m[i][0]==a){
79             ifind=true;
80             break;}
81     }
82     if(ifind)
83         return i;
84     return -1;
85 }
86 void reset(){
87     for (int i=0;i<num;i++)
88     {
89         n[i][0]=m[i][0];
90         n[i][1]=m[i][1];
91         n[i][2]=m[i][2];
92     }
93 }複製代碼