很簡(jiǎn)單的題，給定N個(gè)矩陣，然后給出一系列運(yùn)算式，用括號(hào)來(lái)限制運(yùn)算，最后求一共進(jìn)行了多少次乘法運(yùn)算。
R(m,s) 和Q(s,n)相乘，要進(jìn)行m*s*n次～
處理運(yùn)算先后的時(shí)候，用棧實(shí)現(xiàn)，即后進(jìn)先出就可以了。
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 Outout:
0
0
0
error
10000
error
3500
15000
40500
47500
15125
Code:
#include <iostream>
#include <cstdio>
#include <string>
#include <map>
using namespace std;
struct Matrx{
char id;
int x,y;
}a[30];
Matrx stack[100];
int main()
{
int i,j,k,n,ans;
bool flag;
map<char,int>mark;
scanf("%d",&n);
for(i = 1;i <= n; i++){
cin >> a[i].id;
scanf("%d%d",&a[i].x,&a[i].y);
mark[a[i].id] = i;
}
string str;
while(cin>>str){
ans = 0; flag = true;
int len = str.length();
int head,tail;
head = tail = 0;
for(i = 0;i <len; i++){
if(str[i] == '(')
continue;
else if(str[i] == ')'){ //pop
if(head == 1)
head --;
else if(head > 1){
int tx = stack[head-2].x;
int ty = stack[head-1].y;
if(stack[head-2].y != stack[head-1].x) flag = false;
ans += stack[head-2].x*stack[head-2].y*stack[head-1].y;
head -= 2;
stack[head].x = tx;
stack[head++].y = ty;
}
}
else{ // push
stack[head++] = a[mark[str[i]]];
}
}
if(!flag) printf("error\n");
else printf("%d\n",ans);
}
}