poj1068Parencodings

好厉害好强大，这种思想太厉害了，代码网上找的。

p代表前面”(“的个数，，w代表离“)”最近的右括号的个数

#include<iostream>

#include<string.h>

using namespace std;

int a[101];

int main()

{

int n, m, bf,t;

cin >> t;

while (t--)

{

cin >> n;

bf = 0;

memset(a, 0, sizeof(a));

while (n--)

{

cin >> m;

bf = m;

while (a[m])

{

m -= a[m];

a[bf] += a[m];

}

a[bf]++;

cout << a[bf] << ' ';

}

cout << endl;

}

return 0;

}

在discuss中看到一个精简版的，分析了一下思路：

将P中值相同的分成一类，在每一类集合代表该类中的右括号之前有X个左括号，由于P中数字是递增的，所以X也是递增的。

设立数组a，a[i]=k表示左边有i个左括号的右括号集合已经消耗了从右向左数（因为是就近匹配）的k个括号。

每读入p的一个新数字m，它要么

1. 比上一个大：则意味着产生了新集合，且a[m]=1，并总输出1.

2. 更上一个一样大：意味着在上一个集合上新增一个元素，那么要寻找相匹配的左括号，

首先要跳过该类消耗的括号数a[m]，则还剩余m-a[m]个左括号，这些括号也不是都可以用的

，因为此时m-a[m]集合也有消耗，所以要减去集合消耗的，再往前递推，一直到a[k]=0，表明该处可以匹配。

把记录下跳过的左括号数（即已经被消耗的）+1就是该右括号要输出的结果，同时也要更新a[m]为这个值。

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