POJ 1068 Parencodings-优快云博客

本文链接：https://blog.youkuaiyun.com/chenanzai/article/details/40193967

本文介绍了一种将括号序列通过P-sequence编码为W-sequence的方法，并提供了一个具体的实现示例。该方法对于处理字符串中的括号匹配问题非常有用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;
}