ZOJ Problem Set – 1016 Parencodings

原題：

Parencodings

Time Limit: 1 Second      Memory Limit: 32768 KB

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

Source: Asia 2001, Tehran (Iran)

 

 

雖然是一道簡單題，但是由於是英文的，我做的時候理解起來還是花了那麼一段時間，現在分析一下，主要是理解一下題目，讓我自己以後看的時候不要摸不到頭腦就可以了。

題意是這樣的，現在有一系列的（），要對這些左右圓括弧進行編碼。現在有兩種編碼形式，第一種：從給定的S（左右圓括弧串）中，從左往右說，找到一個右括弧，記錄在這個右括弧之前出現的左括弧的數量，即P編碼。第二種，還是從給定串的左邊開始往右數，找到右括弧，然後從這個位置往左數，記錄與之匹配的第一個左括弧是出現的第幾個左括弧，即W編碼。例如：

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

然後，題目給出了已知S串的P編碼形式，求其W編碼串。雖然根據P編碼形式可以反向求得原S串，但是這卻不是最佳解法。其實我們是可以根據P串直接求W串的。

好啦，題意理解清楚了，就可以給出解題代碼了。

 

  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: }