POJ 1068 -- Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27602		Accepted: 16226

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

 

题意：

对于给出的原括号串，存在两种数字密码串：

1.p序列：当出现匹配括号对时，从该括号对的右括号开始往左数，直到最前面的左括号数，就是pi的值。

2.w序列：当出现匹配括号对时，包含在该括号对中的所有右括号数（包括该括号对）,就是wi的值。

题目的要求：对给出的p数字串，求出对应的w序列。

 

解题思路：

模拟题

给的p序列长度n（1 <= n <= 20）即为右括号的个数。

在处理括号序列时可以使用一个小技巧，把括号序列转化为01序列，左0右1，处理时比较方便而01序列的长度也不会超过40。

 1 #include<iostream>
 2 using namespace std;
 3 int p[21];
 4 int s[41];
 5 int w[21];
 6 int main()
 7 {
 8     int n;
 9     cin>>n;
10     while(n--)
11     {
12         int len;
13         cin>>len;
14         for(int i=0;i<len;i++)
15         {
16             cin>>p[i];
17         }
18         int sj=0;
19         int leftCount = 0;
20         for(int i=0;i<len;i++)
21         {
22             for(int j=0;j<(p[i] - leftCount);j++)
23             {
24                 s[sj] = 0;
25                 sj++;
26             }
27             leftCount = p[i];
28             s[sj] = 1;
29             sj++;
30         }
31         /*
32         //测试s序列代码
33         for(int i=0;i<len*2;i++)
34             cout<<s[i];
35         cout<<endl;
36         */
37         int rightCount = 0;
38         while(rightCount<len)
39         {
40             int temp = 0;
41             for(int i=0;i<len *2;i++)
42             {
43                 if(s[i] == 1) temp++;
44                 if(temp>rightCount)
45                 {///找到当前没有输出的第一个右括号，位置为i
46                     int right = 0;int left = 0;
47                     for(int j=i;j>=0;j--)
48                     {//从第i个位置向前遍历，右括号计数为right，左括号计数为left
49                         //当left == right 时，停止遍历，找到了当前右括号的左括号，进行输出
50                         if(s[j] == 1) right++;
51                         else{
52                             left++;
53                             if(left == right)
54                             {
55                                 cout<<left;
56                                 if(temp != len) cout<<" ";
57                                 else cout<<endl;
58                                 break;
59                             }
60                         }
61                     }
62                     //更新当前找到的右括号的个数
63                     rightCount = temp;
64                 }
65             }
66         }
67     }
68     return 0;
69 }

 

转载于:https://www.cnblogs.com/yxh-amysear/p/8419253.html