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POJ 1068 Parencodings

2014-10-17 23:15 351 查看

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19983		Accepted: 12059

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 <stack>
#include<cstring>
using namespace std;
char a[100];
int b[100];

int f(int n){
stack<char> sta;
int count=0;
for(int i=n;i>=0;i--){
if(a[i]==')'){
sta.push(a[i]);
}
else{
sta.pop();
count++;
}
if(sta.empty()){
return count;
}
}
}
int main(){
int N;
cin>>N;
while(N--){
memset(b,0,sizeof(b));
memset(a,0,sizeof(a));
int n;
cin>>n;
int k,m,count=0;
cin>>m;
k=m;
for(count=0;count<m;count++)
a[count]='(';
a[count++]=')';
k=m;
for(int i=1;i<n;i++){
cin>>m;
for(int j=k;j<m;j++)
a[count++]='(';
a[count++]=')';
k=m;
}
k=0;
for(int i=0;i<=count;i++){
if(a[i]==')')
b[k++]=f(i);
}
for(int i=0;i<n;i++)
cout<<b[i]<<" ";
cout<<endl;
}
return 0;
}

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