Poj OpenJudge 1068 Parencodings

1.Link:

http://poj.org/problem?id=1068

http://bailian.openjudge.cn/practice/1068

2.Content:

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20077		Accepted: 12122

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

Source

Tehran 2001

3.Method:

（1）先将P-sequence转成S序列，方法为：利用vector保存，置入）前，放入当前数值减前一数值的（

（2）将S序列转为W序列，方法为：利用stack，每次遇到（则置入stack，其中-1代表“（”；遇到count = 1，直到遇到第一个（前，将stack的值累加到count，并置出

4.Code:

 #include <iostream>
 #include <vector>
 #include <stack>

 using namespace std;

 int main()
 {
     //freopen("D://input.txt","r",stdin);

     int i,j;

     int t;
     cin >> t;
     while(t--)
     {
         int n;
         cin >> n;

         int *arr_p = new int[n];

         for(i = ; i < n; ++i) cin >> arr_p[i];

         //for(i = 0; i < n; ++i) cout << arr_p[i] << endl;

         vector<char> v_sym;

         int pre_p = ;
         for(i = ; i < n; ++i)
         {
             for(j = pre_p; j < arr_p[i]; ++j) v_sym.push_back('(');
             v_sym.push_back(')');
             pre_p = arr_p[i];
         }

         vector<char>::size_type sym_i;

         //for(sym_i = 0; sym_i != v_sym.size(); ++sym_i) cout << v_sym[sym_i];
         //cout << endl;

         stack<int> s_sym;// -1 (, -2 )
         for(sym_i = ; sym_i != v_sym.size(); ++sym_i)
         {
             if('(' == v_sym[sym_i]) s_sym.push(-);
             else
             {
                 int count = ;
                 while(s_sym.top() != -)
                 {
                     count += s_sym.top();
                     s_sym.pop();
                 }
                 s_sym.pop();
                 cout << count << " ";
                 s_sym.push(count);
             }
         }
         cout << endl;

         delete [] arr_p;

     }

     return ;
 }

5.Reference: