【模拟】POJ-1068-Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27585		Accepted: 16217

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

#include <iostream>
#include<stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[20],w[20];
int vis[40];

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
string s;
memset(w,0,sizeof(w));

int n;
scanf("%d",&n);
for(int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
a[0]=0;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=(a[i]-a[i-1]); j++)
{
s+='(';
}
s+=')';

}

int k=1;
memset(vis,0,sizeof(vis));
for(int i=0; i<2*n; i++)
{
int ans=1;
if(s[i]==')')
{
for(int j=i-1; j>=0; j--)
{
if(s[j]==')')ans++;
if(s[j]=='('&&!vis[j])
{
vis[j]=1;
break;
}
}
w[k++]=ans;
}
}
for(int i=1; i<=n; i++)
{
printf("%d",w[i]);
if(i!=n)printf(" ");
}
printf("\n");

}
}