UVA 442:matrix Chain multiplication Data Structure topic

Topic Link: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_ problem&problem=383

Type of topic: data structure, linked list

Sample input:

9 (
A)
5
D/
F 5
G/5
H I (a)
b
  c
(AA)
(AB)
(AC)
(A (BC))
((AB) C) ((
(((()) ((((
D)) ((((DE) f) (((()) (((()) ((()) ((()) ((( (
(D (EF)) ((GH) I)

Sample output:

0
0
0
error
10000
error
3500
15000
40500
47500
15125

The main effect of the topic:

Given a series of matrices, give them a name of a, B ... And their number of rows and columns. When you're done, you give a series of expressions, and then you ask how many multiplication steps you can take to calculate by these expressions.

This article URL address: http://www.bianceng.cn/Programming/sjjg/201410/45537.htm

The idea of disintegration:

The question is similar to that of the bracket. The key step is to compute this process of matrix multiplication times.

#include <cstdio> #include <cctype> #include <cstring> #include <cstdlib> #include &LT;STACK&G  
T  
    
using namespace Std;  
int sum,n;  
int mat[30][2];  
int arr[30],prev[30], next[30];  
    
Char str[1000];  
    void Solve () {//If there is only one matrix, then direct output 0 if (strlen (str) ==1 && isupper (str[0)) {printf ("0\n");  
        } else{Char copymat[30][2];  
        int i,j; Copy an array to operate.  
            Because the following operations need to modify these matrices for (i=0 i<n; ++i) {copymat[i][0] = mat[i][0];  
        COPYMAT[I][1] = mat[i][1];  
        } sum=0;   
            
        stack<char>st;  
            For (i=0 I<strlen (str); ++i) {if (str[i]== ' (' | | isupper (str[i)) St.push (str[i));        
        else if (str[i]== ') ') {stack<char>temp; When encountering ') ', take out all the matrices in the stack to compute while (Isupper (St.top ())) {Temp.push (St.top ()));  
                St.pop ();  
                }//Put ' (' also pop-up st.pop ();  
                Char ex;  
                    Remove the first matrix (the leftmost one in the original expression) if (!temp.empty ()) {ex=temp.top ();  
                Temp.pop ();  
                    while (!temp.empty ()) {Char multi = temp.top ();  
                    Temp.pop ();  
                        If the number of columns on the left matrix differs from the function of the right matrix, the direct output error if (copymat[ex-' a '][1]!= copymat[multi-' a '][0]) {  
                        printf ("error\n");  
                    return; //Count multiplication, plus sum + = copymat[ex-' A '][0]*copymat[multi-' a '][0]*copymat[multi-'  
                    A '][1];  
                Multiply to get a new matrix, update the size copymat[ex-' a '][1] = copymat[multi-' a '][1];  
            } St.push (ex);    
      }  
        }  printf ("%d\n", sum);  
    int main () {freopen ("Input.txt", "R", stdin);  
    Char ch;  
    int I, J;  
    scanf ("%d%*c", &n);  
    For (i=0 i<n; ++i) {scanf ("%c%d%d%*c", &ch,&mat[i][0],&mat[i][1]);  
    while (gets (str)) {solve ();  
return 0; }