Parencodings
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
這是一道極爲容易的題目，基本原理：根據P排序還原括號字符串S，然後根據W的命名規則查找每一個右括號所對應的編號。僅僅需要簡單的往前回溯即可：假設第i個字符是右括號，則用變量par=1記下當前的右括號數，在往前回溯的過程中，記錄下左括號的個數（leftpar），當par爲0時(此時，左右括號匹配好了)，跳出回溯。此時leftpar爲當前右括號的W編號。具體實現如下：
#include <iostream>
using namespace std;
int main()
{
char S[20];
int P[20], W[20], t, n;
int i, tmp, index, par, leftpar;
cin >> t;
while(t --)
{
cin >> n;
tmp = 0;
index = 0;
for(i = 0; i < n; i ++)
{
cin >> P[i];
if(index < P[i])
{
while(index < P[i])
{
S[tmp++] = '(';
index ++;
}
}
S[tmp ++] = ')';
}
index = 0;
for(i = 0; i < 2*n; i ++)
{
if(S[i] == ')')
{
tmp = i-1;
par = 1;
leftpar = 0;
while(par)
{
if(S[tmp] == ')')
{
par ++;
}
else
{
leftpar ++;
par --;
}
tmp-- ;
}
W[index ++] = leftpar;
}
}
for(i = 0; i < n; i ++)
{
cout << W[i] << " ";
}
cout<<endl;
}
return 0;
}