Uva-442-matrix chain Multiplication

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The key to this question is the parsing expression. The expression in this question is relatively simple and can be completed using a stack --->

When a letter is encountered, it is written into the stack; when there is a right brace, It is computed out of the stack, and then the new matrix is clicked into the stack!

In this case, if the number of columns of A is not equal to the number of rows of B, multiplication cannot be performed !!! Note!

The Code is as follows:

# Include <cstdio> # include <cstring> # include <iostream> # include <stack> # include <string> using namespace STD; struct Ma {int H, L; // horizontal and column! MA (int l = 0, int h = 0): L (L), H (h) {}} M [26]; int main () {int N; cin> N; For (INT I = 0; I <n; I ++) {string name; CIN> name; cin> M [name [0]-'a']. h> M [name [0]-'a']. l;} string SEN; while (CIN> Sen) {stack <Ma> q; int flag = 0, Len = Sen. length (); bool error = false; int ans = 0; For (INT I = 0; I <Len; I ++) {If (isalpha (Sen [I]) // if it is a letter, go to the stack! Q. push (M [Sen [I]-'a']); else if (Sen [I] = ') {Ma m2 = Q. top (); // The first to last! Q. Pop (); MA M1 = Q. Top (); // The second to the last! Q. Pop (); If (m1.h! = M2.l) // if the length of the last 1st columns is not equal to the length of the last 1st columns, exit! {// That is, the horizontal length of! = Column Length of B! Error = true; flag = 1; break;} ans + = m1.h * m1.l * m2.h; q. push (MA (m1.h, m2.l); // Add the new vertex to the stack!} If (FLAG) break; // There is no need to perform cyclic operations!} If (error) cout <"error" <Endl; else cout <ans <Endl;} return 0 ;}