ZOJ Problem Set - 1016 Parencodings

Parencodings

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 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).
• 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 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

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Source: Asia 2001, Tehran (Iran)

題目大意：

把P編碼轉為W編碼。

P編碼的每個數字記錄的是左邊有多少個左括號

W編碼的每個數字記錄的是左邊遇到的和它相匹配的左括號時經過的左括號個數

解題思路：

用堆疊來解題。每一次輸入P的一位數字的時候，用堆疊來模擬括號，其中-1表示左括號，1表示右括號。每次輸入P的一位數字之後，我們將這麼多位左括號（-1）入棧，判斷棧頂為1還是-1，如果為-1則W輸出1，否則持續出棧直到遇到-1，則輸出累加的數字+1。

AC：

#include <cstdio>
#include <string>
#include <stack>
#include <iostream>
using namespace std;

stack<int> S;

int main()
{
	int N;
	cin>>N;
	while(N--)
	{
		int n;
		cin>>n;
		int p; 
		int t = 0;//記錄已經輸入的左括號數（當前P的前序）
		for(int i = 1; i <= n; i++)
		{
			cin>>p;
			for(int j = 1; j <= p - t; j++)
				S.push(-1);			//將左括號全部入棧
			if(S.top() == -1)		//判斷，如果棧頂元素為-1，則輸出1
			{
				i == 1 ? cout<<"1" : cout<<" 1";
				S.pop();		//並出棧-1，入棧1
				S.push(1);
			}
			else                //如果棧首不是-1，則開始計算前面有多少對括號
			{
				int iTop = S.top();//直接出棧棧首
				S.pop();
				while(S.top() != -1)//持續pop，直到遇到-1
				{
					iTop += S.top();
					S.pop();
				}
				cout<<" "<<iTop + 1;
				S.pop();
				S.push(iTop + 1);
			}
			t = p;
		}
		cout<<endl;
		while(!S.empty())
			S.pop();
	}
	system("pause");
	return 0;
}