poj 1068 Parencodings（模拟）

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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;

}

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poj 1068 Parencodings（模拟）

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