Parencodings POJ1068解题报告

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12950 Accepted: 7675

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    Pi=第i个右括号之前的左括号个数
W-sequence    1 1 1456    Wi=第i对匹配的左右括号之间左括号的个数

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

Tehran 2001

思路：先根据P-sequence 还原字符串S（用bool型数组表示），然后利用stack存储每个入栈的左括号的位置j，最后得出结果W-sequence。

代码如下：

#include <iostream>#include <stack>using namespace std;stack<int> s;int main(){freopen("in.txt","r",stdin);int t,n,k,m;int i,j;cin>>t;bool flag[1000]={false};for(i=0;i<t;i++){cin>>n;memset(flag,0,1000);k=0;for(j=0;j<n;j++){cin>>m;flag[k+m]=true;k++;}//for(j=0;j<k+m;j++){cout<<flag[j]<<" ";}//cout<<endl;for(j=0;j<k+m-1;j++){if(!flag[j]){s.push(j);}else{cout<<(j-s.top()+1)/2<<" ";s.pop();}}cout<<(j-s.top()+1)/2<<endl;s.pop();}return 0;}