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 = ;
         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 = ;
                 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 ;
 }