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 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 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;    }}