POJ1068 Parencodings 解题报告

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

分析：

对于给出的序列，明显是一定不会下降的，

用ans[]来记录答案

当A[i]>A[j]时其对应的ans[i]=1;

注意到A[i]-ans[i]其实就是当前右括号对应的左括号的前面有多少个左括号

很明显对每个A[i]-ans[i]都是唯一的,用它来作为匹配过的左括号的编号,这个编号从0开始

所以当A[i]==A[i-1]时 令k=ans[i-1]+1;

不断增加k直到A[i]-k的值没有在前面出现过时,ans[i]=k;

#include <iostream>
#include <cstring>
using namespace std;
int f[100], g[100], ans[100];
int t, n, k, sum;
int main() {
    cin >> t;
    while (t--) {
        cin >> n;
        memset (g, 0, sizeof g);//数组g用来记录使用过的左括号
        for (int i = 1; i <= n; i++) {
            cin >> f[i];
            if (f[i] > f[i - 1]) {//比前面的数大
                ans[i] = 1;//ans[i]赋值为一
                g[f[i] - ans[i]] = 1;//标记使用过的左括号
            }
            else {
                int k = ans[i - 1] + 1;
                while (g[f[i] - k]) k++;//找到没有使用过的左括号
                g[f[i] - k] = 1;
                ans[i] = k;
            }
        }
        cout << ans[1];
        for (int i = 2; i <= n; i++)
            cout << ' ' << ans[i];
        cout << endl;
    }
    return 0;
}

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