POJ 1068, Parencodings

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