Parencodings（POJ 1068）（模拟）

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本文介绍了一个将括号字符串的P序列转换为W序列的程序设计问题。通过输入一个由左括号数量组成的P序列，程序生成对应的W序列，并提供了解决方案的C++实现代码。

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题目链接：http://poj.org/problem?id=1068
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

思考：纯模拟的题不要怕麻烦，按照题中的模拟的步骤做就行了。只要时间复杂度不超就行。

#include <iostream>
#include <cstring>

using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--) {
        int n;
        int a[100];
        char b[1000];
        int vis[1000];
        int ne[1000];
        memset (a, 0, sizeof(a));
        memset (b, '\0', sizeof(b));
        memset (vis, 0, sizeof(vis));
        memset (ne, 0, sizeof(ne));
        cin >> n;
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
        }
        int num = 0;
        int num2 = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < a[i] - a[i - 1]; j++) {
                b[num++] = '(';
            }
            ne[num2++] = num;
            b[num++] = ')';
        }
        for (int i = 0; i < num2; i++) {
            int cnt = 0;
            for (int j = ne[i]; j >= 0; j--) {
                if (b[j] == '(') {
                    cnt++;
                    if (!vis[j]) {
                        vis[j] = 1;
                        break;
                    }
                }
            }
            cout << cnt << " ";
        }
        cout << endl;
    }
    return 0;
}