ZOJ Problem Set – 1016 Parencodings

原题：

Parencodings

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Time Limit: 1 Second      Memory Limit: 32768 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

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Source: Asia 2001, Tehran (Iran)

 

 

虽然是一道简单题，但是由于是英文的，我做的时候理解起来还是花了那么一段时间，现在分析一下，主要是理解一下题目，让我自己以后看的时候不要摸不到头脑就可以了。

题意是这样的，现在有一系列的（），要对这些左右圆括号进行编码。现在有两种编码形式，第一种：从给定的S（左右圆括号串）中，从左往右说，找到一个右括号，记录在这个右括号之前出现的左括号的数量，即P编码。第二种，还是从给定串的左边开始往右数，找到右括号，然后从这个位置往左数，记录与之匹配的第一个左括号是出现的第几个左括号，即W编码。例如：

S (((()()())))
P-sequence 4 5 6666
W-sequence 1 1 1456

然后，题目给出了已知S串的P编码形式，求其W编码串。虽然根据P编码形式可以反向求得原S串，但是这却不是最佳解法。其实我们是可以根据P串直接求W串的。

好啦，题意理解清楚了，就可以给出解题代码了。

 

  1: #include<iostream>
  2: #include<vector>
  3: using namespace std;
  4: 
  5: int main(void)
  6: {
  7:     int times;
  8:     while(cin>>times)
  9:     {
 10:         for(int items;times && cin>>items;times--)
 11:         {
 12:             vector<int> leftParenthsis(items,0);
 13:             int preNumber = 0;
 14:             for(int numbers,i = 0;i < items && cin>>numbers; i++)
 15:             {
 16:                 leftParenthsis[i] = numbers - preNumber;
 17:                 preNumber = numbers;
 18:                 cout<<(i ? " ":"");
 19:                 if(leftParenthsis[i])
 20:                 {
 21:                     cout<<1;
 22:                     leftParenthsis[i]--;
 23:                 }
 24:                 else
 25:                 {
 26:                     for(int j = i; j >= 0;j--)
 27:                     {
 28:                         if(leftParenthsis[j])
 29:                         {
 30:                             cout<<i - j + 1;
 31:                             leftParenthsis[j]--;
 32:                             break;
 33:                         }
 34:                     }
 35:                 }            
 36:             }
 37:             cout<<endl;
 38:         }
 39:     }
 40:     return 0;
 41: }