浙大在线评测 1016 Parencodings

Question:

    Let S = s1 s2 ¡­ s2n be a well-formed string of parentheses. S can be encoded in two different ways:

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). 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 ¡Ü t ¡Ü 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 ¡Ü n ¡Ü 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 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

Solution:

// 声明：本代码仅供学习之用，请不要作为个人的成绩提交。
// http://blog.csdn.net/mskia
// email: [email protected]
#include <iostream>
using namespace std;
int main( void ) {
    int cases;
    cin >> cases;
   
    for ( int i = 0 ; i < cases; ++i ) {
        int n;
        cin >> n;
       
        char* par = new char[ n + n ];
        int*  pos = new int[ n ];
        int lastc = 0 , c = 0 , count1 = 0;
        for ( int j = 0; j < n; ++j ) {
            cin >> c;
            for ( int k = 0; k < c - lastc; ++k ) {
                par[ count1 ] = '(';
                ++count1;
            }
            par[ count1 ] = ')';
            pos[ j ] = count1;
            ++count1;
            lastc = c;
        }
       
        for ( int j = 0; j < n; ++j ) {
            if(j != 0) {
                cout << " ";
            }

            int countleft = 0 , countright = 0;
            for ( int k = pos[ j ]; k >= 0; --k ) {
                switch ( par[ k ] ) {
                    case '(':
                        if ( ++countleft == countright ) {
                            cout << countright;
                            goto NEXTFORJ;
                        }
                        break;
                       
                    case ')':
                        ++countright;
                        break;
                }
            }
            cout << countright;
           
            NEXTFORJ:;
        }
        cout << endl;
       
        delete[] par;
        delete[] pos;
    }
    return 0;
}

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