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轉載請註明出處：http://blog.csdn.net/u012860063?viewmode=contents

題目鏈接：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

題意：

一組括號 (((( ) ( ) ( ) ) ) )

有兩種描寫敘述方法：

P方法：4 5 6 6 6 6 - 每個‘）’前，有幾個‘（’

W方法：1 1 1 4 5 6 - 每個‘)‘的前面第幾個（註意是從當前位置往前面數）’（‘是跟它匹配的

要求是依據P求W

思路：

先依據P還原出括號的位置，在計算出W就可以。

代碼例如以下：

#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（模擬）