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2013
11-09

POJ 1068 Parencodings [解题报告] Java

Parencodings

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

2
6
4 5 6 6 6 6
9 
4 6 6 6 6 8 9 9 9

样例输出:

1 1 1 4 5 6
1 1 2 4 5 1 1 3 9

解题代码:

import java.util.Scanner;

 public class Main {

     int t;
     int len;
     int[] p;
     int[] w;
     String[] s;
     int index;
     int temp;
     int pos;

     public Main() {
         Scanner scan = new Scanner(System.in);
         t = scan.nextInt();
         for (int i = 0; i < t; i++) {
             len = scan.nextInt();
             p = new int[len];
             w = new int[len];
             s = new String[len * 2];
             for (int j = 0; j < len; j++) {
                 p[j] = scan.nextInt();
             }
             pos = 0;
             temp = 0;
             index = 0;
             for (int k = 0; k < len; k++) {
                 for (; pos < p[k]; pos++) {
                     s[index++] = "(";
                 }
                 s[index++] = ")";
             }
             search();
             for (int m = 0; m < len; m++) {
                 System.out.print(w[m] + " ");
             }
             System.out.println();
         }
     }

     public void search() {
         pos = 0;
         int k = 0;
         for (int i = 0; i < len; i++) {
             for (; k < 2 * len; k++) {
                 if (s[k].equals(")")) {
                     pos = k;
                     k++;
                     break;
                 }
             }
             temp = 0;
             index = 1;
             for (int j = pos - 1;; j--) {
                 if (s[j].equals("(")) {
                     temp++;
                     index--;
                 } else {
                     index++;
                 }
                 if (index == 0) {
                     w[i] = temp;
                     break;
                 }
             }
         }
     }

     public static void main(String[] args) {
         new Main();
     } 
}