BZOI 1507 [NOI2003] Editor
Background
After trying to solve problem EDIT1(Editor) and being ****ed by Brainf**k, Blue Mary decided to set another difficult problem about editor.
Description
Some definations:
- Text: It's a sequence that consists characters whose ASCII code is in [32,126].
- Cursor: It's a sign for pointing out the current position.It can be at the start or the end of the text or between two consecutive characters of the text.
Editor is a structure.It contains one text and one cursor.The operations are listed below:
--------------------------------------------------------------------------
| Name | Input format | function |
--------------------------------------------------------------------------
| Move(k) | Move k | Move the cursor after the kth character |
| | | in the text. If k=0, you should put |
| | | the cursor at the start of the text. |
--------------------------------------------------------------------------
| Insert(n,s) | Insert n s | Insert string s whose length is n(>=1) |
| | | after the cursor.The cursor doesn't move. |
--------------------------------------------------------------------------
| Delete(n) | Delete n | Delete n(>=1) characters after the cursor.|
| | | The cursor doesn't move. |
--------------------------------------------------------------------------
| Get(n) | Get n | Output n(>=1) characters after the cursor.|
--------------------------------------------------------------------------
| Prev() | Prev | Move the cursor one character forward. |
--------------------------------------------------------------------------
| Next() | Next | Move the cursor one character backward. |
--------------------------------------------------------------------------
If the text of a editor is empty,we say the editor is empty.
Here is an example._ denotes to the cursor,$ denotes to the start and the end.At start the editor is empty.
------------------------------------------------------------------------------
| Operation | Text after the operation | Output |
------------------------------------------------------------------------------
| INSERT(13,"Balanced tree") | $_Balanced tree$ | $$ |
------------------------------------------------------------------------------
| MOVE(2) | $Ba_lanced tree$ | $$ |
------------------------------------------------------------------------------
| DELETE(5) | $Ba_d tree$ | $$ |
------------------------------------------------------------------------------
| NEXT() | $Bad_ tree$ | $$ |
------------------------------------------------------------------------------
| INSERT(7," editor") | $Bad_ editor tree$ | $$ |
------------------------------------------------------------------------------
| MOVE(0) | $_Bad editor tree$ | $$ |
------------------------------------------------------------------------------
| GET(15) | $_Bad editor tree$ | $Bad editor tree$ |
------------------------------------------------------------------------------
Your task is:
- Build an empty editor.
- Read some operations from the standard input and operate them.
- For each Get operation, write the answer to the output.
Input
the very first line contains the number of testcases T(T<=4).T tests follow.
For each test, the first line is the number of operations N.N operations follow.
Blue Mary is so depressed with the problem EDIT1 that she decides to make the problem more difficult.So she inserts many extra line breaks in the string of the Insert operation.You must ignore them.
Except line breaks, all the charaters' ASCII code are in [32,126]. There's no extra space at the end of a line.
You can assume that for each test case:
- No invalid operation is in the input.
- Number of move operations is no more than 50000.
- Number of the total of insert and delete operations is no more than 4000.
- Number of the total of prev and next operations is no more than 200000.
- The characters inserted will not more than 2MB.The valid output will not more than 3MB.
Output
The output should contain T blocks corresponding to each testcase.
For each test case, the output should contain as many lines as the get operations in the input.Each line should contains the output of each get operation.
Example
Input: 1
15
Insert 26
abcdefghijklmnop
qrstuv wxy
Move 15
Delete 11
Move 5
Insert 1
^
Next
Insert 1
_
Next
Next
Insert 4
.\/.
Get 4
Prev
Insert 1
^
Move 0
Get 22 Output:
.\/.
abcde^_^f.\/.ghijklmno
Warning: large Input/Output data, be careful with certain languages
Blue Mary's note: the test case #1 has something wrong and it has been fixed on April 27th, 2007.Solutions has been rejudged. Please accept my apology.
题目取自SPOJ
几点注意的：
1、bzoj样例有误。
2、Insert操作如果读入长度用scanf("%d\n",&x)读，会自动过滤下一行空格，导致Wa90.
相信这是我写过最差的程序了。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstdlib>
using namespace std;
#define MAXN MAXT
#define MAXT 1024*1024*4+1000
int n,m;
struct Splay_tree
{
struct node
{
node *ch[],*fa;
char w;
int siz;
};
node E[MAXN],nil_node;
queue<node* > Q;
node *root,*nil;
Splay_tree()
{
int i;
for (i=;i<MAXT;i++)
{
Q.push(&E[i]);
}
nil_node.ch[]=nil_node.ch[]=NULL;
nil_node.w='#';
nil_node.siz=;
nil=&nil_node;
root=nil;
root->fa=nil;
}
void update(node *now)
{
if (now==nil)throw "illegal update";
now->siz=now->ch[]->siz+now->ch[]->siz+;
}
void rotate(node *now,int p)//注意不要改变nil的值,now的祖父节点或儿子节点可能为nil
{
node *pnt=now->fa;
now->fa=pnt->fa;
if (pnt->fa!=nil)
{
if (pnt->fa->ch[]==pnt)
{
pnt->fa->ch[]=now;
}else
{
pnt->fa->ch[]=now;
}
}
pnt->fa=now;
pnt->ch[p^]=now->ch[p];
if (now->ch[p]!=nil)now->ch[p]->fa=pnt;
now->ch[p]=pnt;
update(pnt);//注意顺序
update(now);
}
void splay(node *now,node *top)
{
node *pnt;
if (now==top)return ;
while (now->fa!=top)
{
pnt=now->fa;
if (pnt->ch[]==now)
{
if (pnt->fa!=top&&pnt->fa->ch[]==pnt)
{
rotate(pnt,);
}
rotate(now,);
}else
{
if (pnt->fa!=top&&pnt->fa->ch[]==pnt)
{
rotate(pnt,);
}
rotate(now,);
}
}
if (top==nil)
{
root=now;
}
}
node *get_node(int rank)
{
node *now=root;
if (now->siz<rank)throw "Not enough node";
while (true)
{
if (now->ch[]->siz+==rank)
{
return now;
}
if (now->ch[]->siz+<rank)
{
rank-=now->ch[]->siz+;
now=now->ch[];
}else
{
now=now->ch[];
}
}
return now;
}
node *get_min_node(node *now)
{
if (now==nil)throw "illegal call";
//if (now==nil)return nil;
while (now->ch[]!=nil)
{
now=now->ch[];
}
return now;
}
pair<node*,node*> split(int pos)
{
if (pos==)return make_pair(nil,root);
splay(get_node(pos),nil);
pair<node*,node*> ret;
ret.first=root;
ret.second=root->ch[];
root->ch[]->fa=nil;
root->ch[]=nil;
update(root);
root=NULL;
return ret;
}
node * merge(node * a1,node *a2)
{
if (a1==nil)return a2;
if (a2==nil)return a1;
root=a2;
splay(get_min_node(a2),nil);
root->ch[]=a1;
a1->fa=root;
update(root);
return root;
}
void insert(int pos,char ch)//插入ch后前面有pos个字符
{
node *now=Q.front();
Q.pop();
now->w=ch;
if (pos==)
{
if (root==nil)
{
now->fa=nil;
now->ch[]=now->ch[]=nil;
now->siz=;
root=now;
return ;
}
splay(get_min_node(root),nil);
now->fa=root;
root->ch[]=now;
now->ch[]=now->ch[]=nil;
update(now);
update(root);
return ;
}
splay(get_node(pos),nil);
if (root->ch[]!=nil)splay(get_min_node(root->ch[]),root);
now->fa=root;
now->ch[]=root->ch[];
root->ch[]->fa=now;
root->ch[]=now;
now->ch[]=nil;
update(now);
update(root);
splay(now,nil);
return ;
}
void insert2(int pos,char* str,int len)
{
if (len==)return ;
pair<node*,node*> pr1;
pr1=split(pos);
//scan(pr1.first);cout<<endl;
//scan(pr1.second);cout<<endl;
node *now=build_tree(str,,len-,nil);
root=merge(pr1.first,merge(now,pr1.second));
}
node *build_tree(char *str,int l,int r,node *fa)
{
if (l>r)return nil;
node *now=Q.front();
int mid=(l+r)/;
Q.pop();
now->fa=fa;
now->w=str[mid];
now->ch[]=build_tree(str,l,mid-,now);
now->ch[]=build_tree(str,mid+,r,now);
update(now);
return now;
}
void recycle(node *now)
{
if (now==nil)return ;
if (now->fa!=nil)
{
if (now==now->fa->ch[])
{
now->fa->ch[]=nil;
}else if (now==now->fa->ch[])
{
now->fa->ch[]=nil;
}
}
recycle(now->ch[]);
recycle(now->ch[]);
Q.push(now);
} void erase(int pos,int len)
{
pair<node*,node *> pr1,pr2;
pr1=split(pos);
root=pr1.second;
pr2=split(len);
recycle(pr2.first);
root=merge(pr1.first,pr2.second);
}
void scan(node *now)
{
if (now==nil)return;
if (now->siz!=now->ch[]->siz+now->ch[]->siz+)
{
throw "Size error";
}
if (now->ch[]!=nil&&now->ch[]->fa!=now)throw "Wrong ptr";
if (now->ch[]!=nil&&now->ch[]->fa!=now)throw "Wrong ptr";
scan(now->ch[]);
printf("%c",now->w);
scan(now->ch[]);
}
void scan2(node *now)
{
if (now==nil)return;
scan2(now->ch[]);
printf("%c",now->w);
scan2(now->ch[]);
}
void print_str(int pos,int len)
{
if (!len){puts("");return ;}/**/
if (pos==)
{
if (len==root->siz)
{
scan(root);
puts("");
return ;
}
splay(get_node(len+),nil);
scan(root->ch[]);
puts("");
return ;
}
splay(get_node(pos),nil);
if (pos+len<=root->siz)splay(get_node(pos+len),root);
node *temp=root->ch[],*temp2=root->ch[]->ch[];
root->ch[]=nil;root->ch[]->ch[]=nil;
scan2(root->ch[]);puts("");
root->ch[]=temp;root->ch[]->ch[]=temp2;
return ;
}
}spt;
int i;
char str1[MAXN];
int main()
{
//freopen("editor2.in","r",stdin);
//freopen("out2.txt","w",stdout);
try
{
int j,k,x,y;
scanf("%d",&n);
char od[];
int m,p;
char *ptr;
char ch;
int nowad=;
int root2;
for (i=;i<n;i++)
{
scanf("%s ",od);
switch (od[])
{ case 'M': scanf("%d\n",&x);
nowad=x;
break;
case 'I':
scanf("%d",&x);
getchar();
for (j=;j<x;j++)
{
ch=getchar();
if (ch<||ch>)
{
j--;
continue;
}
str1[j]=ch;
//if (i!=1)cerr<<ch<<endl;;
//spt.insert(nowad+j,ch);
}
str1[x]='\0';
spt.insert2(nowad,str1,x);
if (x)ch=getchar();
break;
case 'D':
scanf("%d\n",&x);
spt.erase(nowad,x);
break;
case'G':
scanf("%d\n",&x);
spt.print_str(nowad,x);
break;
case'P':
nowad--;
break;
case'N':
nowad++;
break;
}
/* cout<<od<<" "<<x<<endl;
if (od[0]=='I')cout<<str1<<endl;
cout<<"<<";spt.scan(spt.root);cout<<"["<<nowad<<"]";
cout<<endl;*/
}
}catch (const char* err)
{
cout<<err;
return ;
}
return ;
}
BZOI 1507 [NOI2003] Editor的更多相关文章
- 1507: [NOI2003]Editor（块状链表）
1507: [NOI2003]Editor Time Limit: 5 Sec Memory Limit: 162 MBSubmit: 4157 Solved: 1677[Submit][Stat ...
- 1507: [NOI2003]Editor
1507: [NOI2003]Editor Time Limit: 5 Sec Memory Limit: 162 MB Submit: 3535 Solved: 1435 [Submit][St ...
- 【BZOJ】1507: [NOI2003]Editor（Splay）
http://www.lydsy.com/JudgeOnline/problem.php?id=1507 当练splay模板了,发现wjmzbmr的splay写得异常简介,学习了.orzzzzzzzz ...
- BZOJ 1507 [NOI2003]Editor
Description Input 输 入文件editor.in的第一行是指令条数t,以下是需要执行的t个操作.其中: 为了使输入文件便于阅读,Insert操作的字符串中可能会插入一些回车符,请忽略掉 ...
- BZOJ 1507 NOI2003 Editor Splay
题目大意: 1.将光标移动到某一位置 2.在光标后插入一段字符串 3.删除光标后的一段字符 4.输出光标后的一段字符 5.光标-- 6.光标++ 和1269非常像的一道题,只是弱多了 几个问题须要注意 ...
- BZOJ1507 [NOI2003]Editor 【splay】
1507: [NOI2003]Editor Time Limit: 5 Sec Memory Limit: 162 MB Submit: 4129 Solved: 1660 [Submit][St ...
- BZOJ_1269&&1507_[AHOI2006]文本编辑器editor&&[NOI2003]Editor
BZOJ_1269&&1507_[AHOI2006]文本编辑器editor&&[NOI2003]Editor 题意: 分析: splay模拟即可 注意1507的读入格式 ...
- 【bzoj1507】[NOI2003]Editor /【bzoj1269】[AHOI2006]文本编辑器editor Splay
[bzoj1507][NOI2003]Editor 题目描述 输入 输入文件editor.in的第一行是指令条数t,以下是需要执行的t个操作.其中: 为了使输入文件便于阅读,Insert操作的字符串中 ...
- BZOJ1507 [NOI2003]Editor
是一道裸的Splay(反正我不会Splay,快嘲笑我!) 只需要维护在数列上加点删点操作即可,不会写Splay的渣渣只好Orz iwtwiioi大神一下了.(后来发现程序直接抄来了...) 就当我的第 ...
随机推荐
- [Javascript] IIFE
Javascript modules are a design pattern that allow you to encapsulate your code into smaller self ma ...
- getline和get的区别
#include<iostream> #include<fstream> #include<cstring> using namespace std; int ma ...
- 使用Java辅助类(CountDownLatch、CyclicBarrier、Semaphore)并发编程
在java 1.5中,提供了一些非常有用的辅助类来帮助我们进行并发编程,比如CountDownLatch,CyclicBarrier和Semaphore,今天我们就来学习一下这三个辅助类的用法 一.C ...
- myeclipse 编码问题
在使用eclipse+MyEclipse开发中,许多文件编码默认是ISO-8859-1,不支持中文(如常用的JSP),这样我们每次建文件都要手动改编码,其实我们可以在设置文件默认编码,今后再创建时就不 ...
- Java基础知识强化之IO流笔记26：FileInputStream / FileOutputStream 复制mp4视频的案例
1. 需求:把D:\\English.mp4 复制到当前项目目录下copy.mp4 代码示例: package com.himi.filecopy; import java.io.FileInput ...
- 移动设备日期选择插件（基于JQUERY）
上周花了2个小时写的一个日期选择插件,比较适合移动端的设备.先看个效果图吧.如果刚好是你需要的就往下吧,不需要的也可以继续..... 其实网络上已经有的了类似的成熟插件,比如基于mobiscroll, ...
- svn出错问题（用户名密码有修改以及资源url改变时）
用eclipse 同步SVN服务器宛然无法访问了: org.tigris.subversion.javahl.ClientException: RA layer request failed svn: ...
- 滚动效果，marquee便签
语法: <marquee></marquee> 例子: <marquee><font size=+3 color=red>Hello, World< ...
- Attribute 特性
转载 不错 摘要:纠结地说,这应该算是一篇关于Attribute 的笔记,其中的一些思路和代码借鉴了他人的文笔(见本文底部链接).但是,由于此文对Attribute 的讲解实在是叫好(自夸一下 ...
- Open judge 06月度开销
06:月度开销 总时间限制: 1000ms 内存限制: 65536kB 传送门 描述 农夫约翰是一个精明的会计师.他意识到自己可能没有足够的钱来维持农场的运转了.他计算出并记录下了接下来 N (1 ≤ ...