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  • 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;
    }


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  • 原文地址:https://www.cnblogs.com/asuran/p/1575563.html
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