poj1068 Parencodings

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

解题思路：
其实这是一道模拟题，像这种括号匹配问题，我觉得经常会用到堆，主要就是把各种情况都考虑到，就没有什么问题了，很早之前做的了

代码：
#include<iostream>
#include<stack>
#include<cstring>
using namespace std;
int p[21];
int w[21];
char s[21];
int n;

int p_s()
{
int help;
for(int i = 0;i<n;i++)
{
if(i==0)
{
for(int j = 0;j<p[0];j++)
s[j] = '(';
help = p[0];
s[help] = ')';
}
else if((p[i]-p[i-1])!=0)
{
int help2 = p[i]-p[i-1];
for(int j = 0;j<help2;j++)
{
s[help+1+j] = '(';
}

help = help+help2+1;
s[help] = ')';
}
else if((p[i]-p[i-1])==0)
{
help++;
s[help] = ')';
}
}
//for(int i = 0;i<=help;i++)
// cout<<s[i];
//cout<<endl;
return help;
}

int main()
{
int times ;
cin>>times;
while(times--)
{
int qingkuang = 0;
int m = 0;
cin>>n;
for(int i = 0;i<n;i++)
cin>>p[i];
int big = p_s();
int jilu[200];
memset(jilu,0,sizeof(jilu));
stack<int>sta;
for(int i = 0;i<=big;i++)
{
if(s[i]=='(')
{
sta.push(i);
}
else if(s[i]==')')
{
if(s[i-1]=='(')
{
jilu[i] = 1;
sta.pop();
}
else
{
int x = sta.top();
jilu[i] = (i-x+1)/2;
sta.pop();
}

}
}
for(int i = 0;i<=big;i++)
{
if(jilu[i]!=0)
cout<<jilu[i]<<" ";
}
cout<<endl;
}
}