Poj OpenJudge 1068 Parencodings

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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:

Poj OpenJudge 1068 Parencodings

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原文地址：http://www.cnblogs.com/mobileliker/p/4067816.html