Parencodings（模拟）

ZOJ Problem Set - 1016

Parencodings

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

思路：可以利用给出的P-sequence推导出括号序列，然后再算出W-sequence

W-sequence的定义有点晦涩，在次解释一下。

W-sequence：序列W = w1 w2 ... wn，wi含义：括号序列中第i个右括号和与其匹配的左括号之间有wi个右括号（包括第i个右括号自身）。

AC Code：

#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    vector<pair<char, bool> > par;
    int t, n;
    scanf("%d", &t);
    while(t--)
    {
        par.clear();
        scanf("%d", &n);
        int x, y, cnt = 0;
        while(n--)
        {
            scanf("%d", &x);
            y = x - cnt;
            cnt = x;
            while(y--)
                par.push_back(make_pair('(', false));
            par.push_back(make_pair(')', false));
        }
        vector<pair<char, bool> >::iterator i, j;
        bool first = true;
        for(i = par.begin(); i != par.end(); i++)
        {
            if(i->first == ')')
            {
                x = 0;
                for(j = i; ; j--)
                {
                    if(j->first == ')')
                    {
                        j->second = true;
                        x++;
                    }
                    else if(j->first == '('  && !j->second)
                    {
                        j->second = true;
                        break;
                    }
                    else if(j == par.end()) break;
                }
                if(!first)
                    printf(" %d", x);
                else
                {
                    printf("%d", x);
                    first = false;
                }
            }
        }
        putchar('\n');
    }
    return 0;
}