POJ1068(ACM括号模拟)

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

先根据P数组，将括号数组进行还原。再遍历括号数组，给每个右括号确定自己对应的左括号，并统计M数组的值。

代码：

import java.util.Scanner;

//poj1068:括号问题（ACM模拟）
public class Main {
	static int[] data_P;// P数组
	static int[] data_W;// W数组
	static char[] data;// 括号数组
	static int[] flag;// 左括号标记数组

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int casenum = in.nextInt();// 所有case的数量
		int[] len = new int[casenum];

		int[][] case_P = new int[casenum][];
		for (int num = 0; num < casenum; num++) {
			int ind = in.nextInt();// W数组的长度
			len[num] = ind;
			int[] per_P = new int[ind + 1];// P数组长度多1。
			for (int i = 1; i < per_P.length; i++) {
				per_P[i] = in.nextInt();
			}
			case_P[num] = per_P;
		}
		for (int i = 0; i < casenum; i++) {
			// 初始化各数组
			int inx = len[i];
			data_P = case_P[i];
			data_W = new int[inx];
			data = new char[2 * inx];// 括号数组长度为2倍
			flag = new int[2 * inx];// 对应括号数组
			// 调用还原括号数组的函数
			theBracket(0);
			System.out.println();
		}
	}

	// 还原括号数组--index：数组游标
	public static void theBracket(int index) {
		for (int i = 1; i < data_P.length; i++) {
			int temp = data_P[i] - data_P[i - 1];
			for (int j = 0; j < temp; j++) {// 该右括号与上一个右括号之间左括号的数量
				data[index] = '(';
				index++;
			}
			// 还原完左括号最后还原一个右括号
			data[index++] = ')';
		}
		printdata_W();
	}

	// 计算W数组
	public static void printdata_W() {
		int indd = 0;// W数组游标
		for (int i = 0; i < data.length; i++) {
			int count = 0;// W数组某位置的值
			// 如果找到一个右括号，就以此为起点，回溯到0；期间遇到一个右括号，count+1；
			// 遇到一个括号且标志位不为0，将该标志位置1；break将count值放到W数组中，继续向后找。
			if (data[i] == ')') {
				for (int j = i; j >= 0; j--) {
					if (data[j] == ')') {
						count++;
					}
					if (data[j] == '(' && flag[j] == 0) {
						flag[j] = 1;
						break;
					}
				}
				data_W[indd++] = count;
			}
		}
		// 输出W数组
		for (int i = 0; i < data_W.length; i++) {
			if (i == data_W.length - 1) {
				System.out.print(data_W[i]);
			} else {
				System.out.print(data_W[i] + " ");
			}
		}
	}

}

POJ1068(ACM括号模拟)

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