POJ 1068, Parencodings

妖狐艹你老母 2021-12-22 06:45 151阅读 0赞

模拟类

**Description**  
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:  
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).  
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).

Following is an example of the above encodings:

S  (((()()())))

P-sequence     4 5 6666

W-sequence     1 1 1456

Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.

**Input**  
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.

**Output**  
The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.

**Sample Input**  
2  
6  
4 5 6 6 6 6  
9  
4 6 6 6 6 8 9 9 9

**Sample Output**  
1 1 1 4 5 6  
1 1 2 4 5 1 1 3 9

**Source**  
Tehran 2001

--------------------

//  POJ1068.cpp : Defines the entry point for the console application.  
 //  
 \#include  < iostream >   
\#include  < string >   
\#include  < iterator >   
\#include  < vector >   
 using   namespace  std;  
  
 int  main( int  argc,  char \*  argv\[\])  
\{  
     int  cases;  
     int  parentheses;  
     int  brackets;  
    cin  >>  cases;  
    vector < bool >  line;  
    vector < int >  wseq;  
     for ( int  i  =   0 ; i  <  cases ;  \++ i)  
    \{  
        line.clear();  
        wseq.clear();  
        cin  >>  parentheses;  
         int  k  =   0 ;  
         for  ( int  j  =   0 ; j  <  parentheses;  \++ j)  
        \{  
            cin  >>  brackets;  
              
             while  (k  <  brackets)  
            \{  
                 \++ k;  
                line.push\_back( true );  
            \}  
  
            line.push\_back( false );  
        \}  
  
         for  ( int  j  =   0 ; j  <  line.size();  \++ j)  
        \{  
             if  (line\[j\]  ==   false )  
            \{  
                 int  match  =   \- 1 ;  
                 int  k  =  j;  
                 int  cnt  =   0 ;  
                 while  (match  !=   0 )  
                \{  
                     \-- k;  
                     if  (line\[k\]  ==   true )  \++ match,  \++ cnt;  
                     else   \-- match;  
                \}  
                wseq.push\_back(cnt);  
            \}  
        \}  
        copy(wseq.begin(),wseq.end(),ostream\_iterator < int > (cout, "   " ));  
        cout  <<   " \\n " ;  
    \}  
     return   0 ;  
\}

转载于:https://www.cnblogs.com/asuran/archive/2009/09/28/1575563.html