POJ Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9010		Accepted: 5321

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

我的理解：

      明白括号串两种表示方法的具体内容，通过"栈"合理将第一种表示方法转向第二种表示方法。

我的代码：

/**
 *
 * POJ 1068 Parencodings
 * 括号匹配问题. 已知描述括号串的两种问题。
 * 方法1：对应每个右括号前面左括号的个数
 * 方法2：相当于每个右括号，内层的左括号的个数
 */

#include <iostream>
#include <vector>
#include <string> 
#include <fstream>            

using namespace std;

string str="";                 // 记录括号序列

// 计算右括号内层的左括号数
int count( int index )
{
 int i;
 int cnt=0;
 for( i=index-1; i>=0; i-- )
 {
  if( str[i]=='(' )                  // 如果遇到左括号则与当前右括号匹配
  {
   str.erase( str.begin() + i );  // 删除当前左括号
   break;
  }
  else if( str[i]==')' )             // 右括号
  {
   cnt++;                         // 计数器叠加
  }
 }

 return cnt+1;                          // 返回
}

// 求解函数
void function()
{
 int i;
 bool flag = false;
 for( i=0; i<str.length();)             // 遍历当前括号串
 { 
  if( str[i]==')' )                  // 遇到右括号输出相应结果
  {
   if( flag==true)
    cout<<" ";
   cout<<count(i);                //
   flag=true;
  } 
  else
  {
   i++;
  }
 }

 cout<<endl;
}

int main()
{
// ifstream cin("test.txt");

 int n;                     // 测试数据的组数
 int m;                     // 描述括号状态的整数
 int state;                 // 括号状态描述变量
 int i,j;                   // 循环变量

 cin>>n;                    // 获取测试数据数

 while( n-- )
 {
  str="";

  cin>>m;                // 获取m

  int last=0;            // 记录上一个括号状态量
  // 生成括号串
  for( i=0; i<m; i++ )
  {
   cin >>  state;     // 获取state

   j=0;
   while( j< (state-last) )
   {
    str+="(";
    j++;
   }
   last = state;

   str+=")";
  }

  function();
 }

 return 0;
}