【dfs】Sequence Decoding

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.

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【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。

 #include<bits/stdc++.h>
 using namespace std;
 const int N = 1e5+;
 char s[N];
 int a[N],n,T;
 void dfs( int L , int R ){
     int num =  ;
     for( int i = L ; i <= R ;i ++ ){
         if( s[i] == 'P' || s[i] == 'H' ){
             printf("%c",s[i] );
         }else if( isdigit(s[i]) ){
             int j = i ;
             num =  ;
             while( isdigit(s[j]) && j < n ){
                 num = num *  + s[i] - '' ;
                 j ++ ;
             }
             i = j -  ;
         }else if( s[i] == '[' ){
             for( int j =  ; j < num ; j++ )
                 dfs( i+ , a[i] -  );
             num =  ;
             i = a[i] ;
         }else if( s[i] == ']' ){
             continue ;
         }
     }
 }

 int main()
 {
     ios_base :: sync_with_stdio(false);
     cin.tie(NULL) , cout.tie(NULL) ;
     cin >> T ;

     while(T--){
         cin >> s;
         n =  strlen( s );
         stack<int> st;
         for( int i =  ; i < n ; i++ ){
             if( s[i] == '[' ){
                 st.push(i) ;
             }else if( s[i] == ']' ){
                 int cur = st.top() ;
                 a[cur] = i ;
                 st.pop();
             }
         }
         dfs(  , n- );
         puts("")    ;
     }
     return ;
 }