Parencodings 1068

最新推荐文章于 2022-03-13 15:34:27 发布

原创最新推荐文章于 2022-03-13 15:34:27 发布 · 191 阅读

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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.

思路：source为输入，delta为匹配了该右括号之后，该右括号之前还有多少个左括号没匹配，result保存结果。

若delta[i]>=0，说明该距离上个右括号之前就有未匹配的左括号，故result[i]=0

若delta[i] < 0,说明要往前遍历，找到第一个delta不为0的右括号。

#include<iostream>
using namespace std;
int source[21];
int delta[21];
int result[21];
int main(){
    int total_case;
    cin>>total_case;
    source[0] = 0;
    while(total_case--){
        int total_nums,j;
        cin>>total_nums;
        for(int i = 1; i <= total_nums;i++){
            cin>>source[i];
            delta[i] = source[i]-source[i-1]-1;
        }
        for(int i = 1; i <= total_nums;i++){

            if(delta[i] == -1){
                for(j = i-1; j >= 0; j--){
                    if(delta[j] > 0)
                        break;
                }
                result[i] = i-j+1;
                delta[j]--;
            }else{
                result[i] = 1;
            }
        }
        for(int i = 1; i <= total_nums;i++){
            cout<<result[i]<<" ";
        }
        cout<<endl;
    }
}