标签：acid show splay space integer pes wing job out
Sequence Decoding
题目描述
The amino acids in proteins are classified into two types of elements, hydrophobic (nonpolar) and hydrophilic (polar). Hydrophobic and hydrophilic are denoted by H and P respectively. A protein is represented by a sequence of H and P such as PPHHHPHPH. In order to reduce the representation length of a sequence, we use the notation k[S] to denote the repeated sequence of k times sequence S,where 2 ≤ k ≤ 9. A legal sequence is defined as following.
·A sequence consists of only one character ‘H’ or ‘P’ is a legal sequence.
·Let S1 and S2 be legal sequences. Then the sequence concatenated by S1 and S2 is also a legal sequence.
·Let S be a legal sequence. Then the sequence k[S] is also a legal sequence, where 2 ≤ k ≤ 9.
For example, PHPHPHPH is encoded as 4[PH]. Note that a repeated sequence may contains repeated sequences recursively such as 2[PH4[P]4[H]].
Given a nonempty encoded protein sequence S, your job is to expand S to its original sequence.
That is, you should expand 2[PH4[P]4[H]] to PHPPPPHHHHPHPPPPHHHH.
输入
The first line is an integer n indicating the number of test cases. Each of the next n lines consists of a legal sequence composed by number digits ‘2’~’9’, ‘[’, ‘]’, ‘P’ and ‘H’.
输出
For each test case, output the expanding sequence in one line.
样例输入
3
PHPHP
2[3[P]H2[P]]
HH2[P3[H]]P
样例输出
PHPHP
PPPHPPPPPHPP
HHPHHHPHHHP
提示
·1 ≤ n ≤ 10.
·All the inputs are legal.
·The length of each input sequence is less than 50.
·The length of each expanded sequence is less than 1000.
【题解】：
一开始想着怎么用栈来模拟，后来写不出来，队友提示用dfs写就好了。
我就用dfs写了。
1 #include<bits/stdc++.h>
2 using namespace std;
3 const int N = 1e5+10;
4 char s[N];
5 int a[N],n,T;
6 void dfs( int L , int R ){
7 int num = 0 ;
8 for( int i = L ; i <= R ;i ++ ){
9 if( s[i] == ‘P‘ || s[i] == ‘H‘ ){
10 printf("%c",s[i] );
11 }else if( isdigit(s[i]) ){
12 int j = i ;
13 num = 0 ;
14 while( isdigit(s[j]) && j < n ){
15 num = num * 10 + s[i] - ‘0‘ ;
16 j ++ ;
17 }
18 i = j - 1 ;
19 }else if( s[i] == ‘[‘ ){
20 for( int j = 0 ; j < num ; j++ )
21 dfs( i+1 , a[i] - 1 );
22 num = 1 ;
23 i = a[i] ;
24 }else if( s[i] == ‘]‘ ){
25 continue ;
26 }
27 }
28 }
29
30 int main()
31 {
32 ios_base :: sync_with_stdio(false);
33 cin.tie(NULL) , cout.tie(NULL) ;
34 cin >> T ;
35
36 while(T--){
37 cin >> s;
38 n = strlen( s );
39 stack<int> st;
40 for( int i = 0 ; i < n ; i++ ){
41 if( s[i] == ‘[‘ ){
42 st.push(i) ;
43 }else if( s[i] == ‘]‘ ){
44 int cur = st.top() ;
45 a[cur] = i ;
46 st.pop();
47 }
48 }
49 dfs( 0 , n-1 );
50 puts("") ;
51 }
52 return 0;
53 }
View Code
【dfs】Sequence Decoding
标签：acid show splay space integer pes wing job out
原文地址：https://www.cnblogs.com/Osea/p/11569188.html
