Parencodings
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 28491 | Accepted: 16786 |
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
大致题意：
一个括号表达式可以按照如下的规则表示，就是每个右括号之前的左括号数。
比如(((()()())))，每个右括号之前的左括号数序列为P=4 5 6 6 6 6，而每个右括号所在的括号内包含的括号数为W=1 1 1 4 5 6.
现在给定P，输出W。
解题思路：
先根据P还原整个括号表达式，存在数组中，然后递归解出W..
代码：
#include<iostream>
using namespace std;
char y[10000]; int p[50],w[50],n,l,j; //y为还原后的括号
int f()
{
int s=1; //消除之前的连续左括号对结果的影响
while(1)
if(y[j]=='(') //若是左括号，继续递归求解
{
j++;
s+=f();
}
else //若是右括号，将已有括号数写入W
{
w[l++]=s;
j++;
return s;
}
}
int main()
{
int t,i,k;
cin>>t; //例子个数
while(t--)
{
cin>>n; //右括号个数
for(i=0,l=0,k=0;i<n;i++)
{
cin>>p[i]; //输入P
for(j=0;j<p[i]-k;j++) //每个右括号前的左括号加入序列
y[l++]='(';
y[l++]=')'; //加入此次右括号
k=p[i]; //之前加过的左括号去掉，以免重复
}
l=j=0;
f();
for(i=0;i<n;i++)
cout<<w[i]<<' ';
cout<<endl;
}
return 0;
}