【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+10;
char s[N];
int a[N],n,T;
void dfs( int L , int R ){
    int num = 0 ;
    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 = 0 ;
            while( isdigit(s[j]) && j < n ){
                num = num * 10 + s[i] - '0' ;
                j ++ ;
            }
            i = j - 1 ;
        }else if( s[i] == '[' ){
            for( int j = 0 ; j < num ; j++ )
                dfs( i+1 , a[i] - 1 );
            num = 1 ;
            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 = 0 ; i < n ; i++ ){
            if( s[i] == '[' ){
                st.push(i) ;
            }else if( s[i] == ']' ){
                int cur = st.top() ;
                a[cur] = i ;
                st.pop();
            }
        }
        dfs( 0 , n-1 );
        puts("")    ;
    }
    return 0;
}