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:

 1 #include <iostream>

 2 #include <vector>

 3 #include <stack>

 4 

 5 using namespace std;

 6 

 7 int main()

 8 {

 9     //freopen("D://input.txt","r",stdin);

10 

11     int i,j;

12 

13     int t;

14     cin >> t;

15     while(t--)

16     {

17         int n;

18         cin >> n;

19 

20         int *arr_p = new int[n];

21 

22         for(i = 0; i < n; ++i) cin >> arr_p[i];

23 

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

25 

26 

27         vector<char> v_sym;

28 

29         int pre_p = 0;

30         for(i = 0; i < n; ++i)

31         {

32             for(j = pre_p; j < arr_p[i]; ++j) v_sym.push_back('(');

33             v_sym.push_back(')');

34             pre_p = arr_p[i];

35         }

36 

37         vector<char>::size_type sym_i;

38 

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

40         //cout << endl;

41 

42         stack<int> s_sym;// -1 (, -2 )

43         for(sym_i = 0; sym_i != v_sym.size(); ++sym_i)

44         {

45             if('(' == v_sym[sym_i]) s_sym.push(-1);

46             else

47             {

48                 int count = 1;

49                 while(s_sym.top() != -1)

50                 {

51                     count += s_sym.top();

52                     s_sym.pop();

53                 }

54                 s_sym.pop();

55                 cout << count << " ";

56                 s_sym.push(count);

57             }

58         }

59         cout << endl;

60 

61 

62 

63         delete [] arr_p;

64 

65 

66     }

67 

68     return 0;

69 }

 

5.Reference: