POJ1068 STL解法思考

以下是对f_zyj大佬模板中提供的利用string和stack模拟过程的学习和注解

f_zyj大佬的博客:

http://blog.csdn.net/f_zyj/article/details/51594851

地址为pdf中第14页

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 6666W-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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

Source

Tehran 2001

#define pause system("pause");#define fence puts("-------------");#define endline puts("")/*好久没读英文题了,边查边看好无奈..意思大概是给一串括号这个括号根据他的规则有两种表示形式要求 输入第一种 输出第二种p[i]表示括号字符串中每个')'前面有多少个'('w[i]表示括号字符串中每个')'匹配'('成功时,匹配到的'('到')'之间成功匹配的对数,自身算一个如(((()()())))中间三个是1,因为他们中间没有成功()对往外一层一层加一所以是1 1 1 4 5 6//这里看题意看了好久题目规模很小,所以大佬用了 string 和 stack 模拟*/#include <iostream>#include <string>#include <stack>using namespace std;int main(){int n;cin >> n;for (int i = 0; i<n; i++){int m;cin >> m;string str;int leftpa = 0;//辅助记录当前'('的个数for (int j = 0; j<m; j++)  // 读入P编码，构造括号字符串{int p;cin >> p;for (int k = 0; k<p - leftpa; k++) str += '(';str += ')';//每一次循环加一个')'leftpa = p;}stack<int> s;for (string::iterator it = str.begin(); it != str.end(); it++){   // 构造M编码//我对栈结构的理解很浅啊,这么简单的用法都看了半天,需要加强练习!.if (*it == '(')s.push(1);//待匹配else{int p = s.top(); s.pop();//遇到')'开始计算,取出一个用于匹配,那么他前一个就是现在的top就加上了这一个,表示从这个左括号开始有一个成对的括号了.cout << p << " ";if (!s.empty()) s.top() += p;//同理,取出的那个元素的值是多少,就有多少对}}cout << endl;}pause}