cf670E Correct Bracket Sequence Editor
Recently Polycarp started to develop a text editor that works only with correct bracket sequences (abbreviated as CBS).
Note that a bracket sequence is correct if it is possible to get a correct mathematical expression by adding "+"-s and "1"-s to it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not. Each bracket in CBS has a pair. For example, in "(()(()))":
- 1st bracket is paired with 8th,
- 2d bracket is paired with 3d,
- 3d bracket is paired with 2d,
- 4th bracket is paired with 7th,
- 5th bracket is paired with 6th,
- 6th bracket is paired with 5th,
- 7th bracket is paired with 4th,
- 8th bracket is paired with 1st.
Polycarp's editor currently supports only three operations during the use of CBS. The cursor in the editor takes the whole position of one of the brackets (not the position between the brackets!). There are three operations being supported:
- «L» — move the cursor one position to the left,
- «R» — move the cursor one position to the right,
- «D» — delete the bracket in which the cursor is located, delete the bracket it's paired to and all brackets between them (that is, delete a substring between the bracket in which the cursor is located and the one it's paired to).
After the operation "D" the cursor moves to the nearest bracket to the right (of course, among the non-deleted). If there is no such bracket (that is, the suffix of the CBS was deleted), then the cursor moves to the nearest bracket to the left (of course, among the non-deleted).
There are pictures illustrated several usages of operation "D" below.

All incorrect operations (shift cursor over the end of CBS, delete the whole CBS, etc.) are not supported by Polycarp's editor.
Polycarp is very proud of his development, can you implement the functionality of his editor?
The first line contains three positive integers n, m and p (2 ≤ n ≤ 500 000, 1 ≤ m ≤ 500 000, 1 ≤ p ≤ n) — the number of brackets in the correct bracket sequence, the number of operations and the initial position of cursor. Positions in the sequence are numbered from left to right, starting from one. It is guaranteed that n is even.
It is followed by the string of n characters "(" and ")" forming the correct bracket sequence.
Then follow a string of m characters "L", "R" and "D" — a sequence of the operations. Operations are carried out one by one from the first to the last. It is guaranteed that the given operations never move the cursor outside the bracket sequence, as well as the fact that after all operations a bracket sequence will be non-empty.
Print the correct bracket sequence, obtained as a result of applying all operations to the initial sequence.
8 4 5
(())()()
RDLD
()
12 5 3
((()())(()))
RRDLD
(()(()))
8 8 8
(())()()
LLLLLLDD
()()
In the first sample the cursor is initially at position 5. Consider actions of the editor:
- command "R" — the cursor moves to the position 6 on the right;
- command "D" — the deletion of brackets from the position 5 to the position 6. After that CBS takes the form (())(), the cursor is at the position 5;
- command "L" — the cursor moves to the position 4 on the left;
- command "D" — the deletion of brackets from the position 1 to the position 4. After that CBS takes the form (), the cursor is at the position 1.
Thus, the answer is equal to ().
先预处理出每个(对应的),每个)对应的(
这个用个栈就好
维护一个像是双端链表的东西,左右移就直接在链表上移,对于每个删除操作,把它到它对应的符号的位置一段全删掉即可
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,pos;
int go[];
char s[];
char op[];
int zhan[],top;
int d[],l[],r[];
int main()
{
n=read();m=read();pos=read();
scanf("%s",s+);
scanf("%s",op+);
for (int i=;i<=n;i++)
{
if (s[i]=='(')zhan[++top]=i;
else
{
go[i]=zhan[top];
go[zhan[top]]=i;
top--;
}
d[i]=s[i]=='(';
if (i!=)l[i]=i-;
if (i!=n)r[i]=i+;
}
for (int i=;i<=m;i++)
{
if (op[i]=='L'){if (l[pos])pos=l[pos];}
if (op[i]=='R'){if (r[pos])pos=r[pos];}
if (op[i]=='D')
{
int nex=go[pos];
if (nex<pos)swap(nex,pos);
l[r[nex]]=l[pos];
r[l[pos]]=r[nex];
if (r[nex])pos=r[nex];else pos=l[pos];
}
}
while (l[pos])pos=l[pos];
if (!pos){puts("");return ;}
while (r[pos])
{
printf("%c",d[pos]==?'(':')');
pos=r[pos];
}
printf("%c",d[pos]==?'(':')');
}
cf 670E
cf670E Correct Bracket Sequence Editor的更多相关文章
- CodeForces 670E Correct Bracket Sequence Editor(list和迭代器函数模拟)
E. Correct Bracket Sequence Editor time limit per test 2 seconds memory limit per test 256 megabytes ...
- Codeforces Round #350 (Div. 2) E. Correct Bracket Sequence Editor 栈 链表
E. Correct Bracket Sequence Editor 题目连接: http://www.codeforces.com/contest/670/problem/E Description ...
- Codeforces Round #350 (Div. 2) E. Correct Bracket Sequence Editor 线段树模拟
E. Correct Bracket Sequence Editor Recently Polycarp started to develop a text editor that works o ...
- 【31.93%】【codeforces 670E】Correct Bracket Sequence Editor
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- Codeforces Round #350 (Div. 2) E. Correct Bracket Sequence Editor 模拟
题目链接: http://codeforces.com/contest/670/problem/E 题解: 用STL的list和stack模拟的,没想到跑的还挺快. 代码: #include<i ...
- Codeforces Round #350 (Div. 2) E. Correct Bracket Sequence Editor (链表)
题目链接:http://codeforces.com/contest/670/problem/E 给你n长度的括号字符,m个操作,光标初始位置是p,'D'操作表示删除当前光标所在的字符对应的括号字符以 ...
- CodeForces 670E Correct Bracket Sequence Editor
链表,模拟. 写一个双向链表模拟一下过程. #pragma comment(linker, "/STACK:1024000000,1024000000") #include< ...
- Codeforces 670E - Correct Bracket Sequence Editor - [线段树]
题目链接:https://codeforces.com/contest/670/problem/E 题意: 给出一个已经匹配的括号串,给出起始的光标位置(光标总是指向某个括号). 有如下操作: 1.往 ...
- Codeforces 670E - Correct Bracket Sequence Editor - [链表]
题目链接:https://codeforces.com/contest/670/problem/E 题意: 给出一个已经匹配的括号串,给出起始的光标位置(光标总是指向某个括号). 有如下操作: 1.往 ...
随机推荐
- C# 语言 类
++++String类+++++黑色小扳手 - 属性紫色立方体 - 方法 ***字符串.Length - 字符串长度,返回int类型 字符串.TrimStart() - 去掉前空格字符串.TrimEn ...
- php 正则符号说明
preg_match_all ("/<b>(.*)<\/b>/U", $userinfo, $pat_array); preg_match_all (&qu ...
- Python 字符编码问题的处理
python中的字符编码问题往往是初学者容易弄不明白的问题, 要想将这个问题搞清楚,需要先弄明白以下的概念 decode 和 encode 函数的作用 字符串字面量的编码格式 decode(str) ...
- jQuery备忘录
jquery 中遍历数组 var arr = [1,2,3,4,5] $.each(arr,function(i,j){ console.log(i,j) }) 结果 0 1 1 2 .... jQu ...
- 基于Python的Web应用开发实战——3 模板
要想开发出易于维护的程序,关键在于编写形式简洁且结构良好的代码. 当目前为止,你看到的示例都太简单,无法说明这一点,但Flask视图函数的两个完全独立的作用却被融合在了一起,这就产生了一个问题. 视图 ...
- Java中System.setProperty()
Java中System.setProperty()用法 <转抄> // Daysafter :Integer中 getInteger( String s); getInteger( Str ...
- 3d点云
rgb-d:rgb加depth组成4channel的 3d点云
- Codeforces Round #277.5 (Div. 2)-C. Given Length and Sum of Digits...
http://codeforces.com/problemset/problem/489/C C. Given Length and Sum of Digits... time limit per t ...
- UEditor1.4.3的实例程序
官网:http://ueditor.baidu.com/website/ 配置下就可以使用 (1)下载,解压后文件结构如下: (2)将整个文件夹改名ueditor后复制到WebRoot目录下: (3) ...
- shell脚本,awk替换{}里面的内容
如何将oxo{axbxc}oxo{dxexf}oxo里面的{}里面的x 替换为; 用awk实现 [root@localhost 09-30]# echo 'oxo{axbxc}oxo{dxexf}ox ...