ZOJ1015 Parencodings

Parencodings

Time Limit: 2000 msMemory Limit: 65536 KB

Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 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).
• 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 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

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
 
using namespace std;
const int N = 59;
int main()
{
    freopen("in.txt","r",stdin);
    char ch[N];
    int n,p[N],ans[N],pp[N],cas;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&p[i]);
        }
        int cnt = 0;
        for(int i=0,j=0;i<(n<<1);i++)
        {
            if(cnt<p[j])
            {
                ch[i] = '(';
                cnt++;
            }else
            {
                j++;
                ch[i] = ')';
            }
        }

        cnt = 1 ;
        memset(ans,-1,sizeof(ans));
        for(int i=0;i<(n<<1);i++)
        {
            if(ch[i]=='(')
            {
                pp[cnt++] = i;
            }else
            {
                ans[i] = pp[--cnt];
            }
        }
        int ou = 0;
        for(int i=0;i<(n<<1);i++)
        if(ans[i]>=0)
        {
            if(ou) printf(" ");
            else ou = 1;
            printf("%d",(i - ans[i]+1)/2);
        }
        printf("\n");
    }
    return 0;
}