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poj 1068 Parencodings（模拟）

2024-10-15 21:15:40   215人阅读

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题目链接：http://poj.org/problem?

id=1068



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




题意：


一组括号 (((( ) ( ) ( ) ) ) )



有两种描写叙述方法：



P方法：4 5 6 6 6 6 - 每个‘）’前，有几个‘（’



W方法：1 1 1 4 5 6 - 每个‘)‘的前面第几个（注意是从当前位置往前面数）’（‘是跟它匹配的



要求是依据P求W




思路：


先依据P还原出括号的位置，在计算出W就可以。







代码例如以下：




#include <iostream>
using namespace std;
#include <cstring>
#define N 117
char s[10000];
int n;
int i, j, k, l;
int p[N],w[N], flag[N];
void WW()
{
	int x = 1;
	int count = 0;
	for(i = 1; i < l; i++)
	{
		if(s[i] == ‘)‘)
		{
			for(j = i-1; j >= 1; j--)
			{
				if(s[j] == ‘(‘)
				count++;
				if(flag[j] == 0 && s[j] == ‘(‘)
				{
					flag[j] = 1;
					w[x++] = count;
					break;
				}
			}
			count = 0;
		}
	}
}
int main()
{
	int t;	
	while(cin >> t)
	{
		while(t--)
		{
			memset(flag,0,sizeof(flag));
			cin >> n;
			k = 0, l = 1;
			for(i = 1; i <= n; i++)
			{
				cin >> p[i];
				for(j = 1; j <= p[i] - k; j++)
				{
					s[l++] = ‘(‘;
				}
				s[l++] = ‘)‘;
				k = p[i];
			}
		//	cout<<s<<endl;
			WW();
			for(i = 1; i < n; i++)
			{
				cout<<w[i]<<‘ ‘;
			}
			cout<<w[n]<<endl;
		}
	}
	return 0;
}











   

poj 1068 Parencodings（模拟）






























        

            
                
                  
                
                

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