CVS

Description

Yoda: Visit I will the cloners on Kamino... And see this army they have created for the Republic.
Cloners from the Kamino planet breed some of the finest clones. Such good results are due to the careful management over the clones’ evolution. The Kaminuans are now busy working out a new study technology that lets increase the clones’ effectiveness. The cloners have come up with a new control system CVS (Clone Version System) that makes managing the progress of experiments easier. The system is quite simple to use.
The Kaminuans have some set of educational programs at their disposal. A clone’s effectiveness depends on which programs and in which order he has learned. The Kaminuans can teach any clone a program as long as this clone hasn’t learned it already.
To make the experiments even easier to conduct, the Kaminuans enabled canceling the changes made by the last program the clone has learned. In this case, the clone’s knowledge returns to the level when the program hasn’t yet been studied. Then this clone can study this program in the future. You can cancel programs at any time unless the clone is at the basic knowledge level.
Besides the ‘roll back’ function, a clone can ‘re-learn’ a program. If one cancels some program by mistake, he can cancel the cancellation. The CVS keeps record of each clone’s canceled programs. After a program is canceled, the CVS adds another record. After a clone re-learn a program, the record is deleted. If a clone learn (not relearn) a program, all cancellation record history of this clone is deleted. You can use the re-learn function as long as the record history for this clone contains any records.
Finally, the system has a ‘clone’ option. If a Kaminuan likes the current variant of a clone, he can clone the clone, that is, create a new clone with the same sequence of taught programs and cancellation history.
Initially the Kaminuans have a single clone with basic knowledge. Help them analyze the progress of the experiments.

Input

The first line of the input contains numbers n — the number of queries — and m — the number of educational programs (1 ≤ nm ≤ 5·10 5). Each of the following n lines has one of the formats given below.

  • learn ci pi. Teach clone ci program pi (1 ≤ pi ≤ m).
  • rollback ci. Cancel the last learned program for clone ci.
  • relearn ci. Apply ‘re-learn’ function to clone ci.
  • clone ci. Clone the clone ci.
  • check ci. Display the last program clone ci has learned and knows at the moment.

It is guaranteed that rollback won’t be applied to the clone that is at the basic knowledge level. learn is always applied with the program a clone doesn’t already know. relearn is only applied if the cancellation history of a clone is not empty. It is also guaranteed that only the clones that already exist can occur in the queries. The numbers are assigned to the clones in the order the clones appear. The Kaminuans started their experiments from clone number one.

Output

For each check  ci query display the result on a single line. If some clone has only basic knowledge, print basic, otherwise print the number of the last learned program.

Sample Input

input output
9 10
learn 1 5
learn 1 7
rollback 1
check 1
clone 1
relearn 2
check 2
rollback 1
check 1
5
7
basic
 
题目值得一做
解析:
通过两个线性表来实现前进后退操作。
 
#include <cstdio>
#include <iostream>
#define N 500005
using namespace std;
int now[N],tot_ro,c,p;
int stack[N],pre[N],tot;
int stack_ne[N],now_ne[N],next[N],tot_ne;
char order[];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
tot=tot_ne=;
tot_ro=;
now[]=now_ne[]=;
for(int i=; i<n; i++){
scanf("%s",order);
switch (order[]) {
case 'n'://learn
scanf("%d%d",&c,&p);
stack[++tot]=p;
pre[tot]=now[c];
now[c]=tot;
now_ne[c]=;
break;
case 'b'://rollback
scanf("%d",&c);
if(now[c]!=){
stack_ne[++tot_ne]=stack[now[c]];
next[tot_ne]=now_ne[c];
now_ne[c]=tot_ne;
now[c]=pre[now[c]];
}
break;
case 'a'://relearn
scanf("%d",&c);
if(now_ne[c]!=){
stack[++tot]=stack_ne[now_ne[c]];
pre[tot]=now[c];
now[c]=tot;
now_ne[c]=next[now_ne[c]];
}
break;
case 'e'://clone
scanf("%d",&c);
if(c<=tot_ro){
now[++tot_ro]=now[c];
now_ne[tot_ro]=now_ne[c];
}
break;
case 'k'://check
scanf("%d",&c);
if(now[c]==)
printf("basic\n");
else
printf("%d\n",stack[now[c]]);
break;
}
}
}
return ;
}

URAL 1992的更多相关文章

  1. URAL 1992 CVS

    CVS 题目连接: http://acm.timus.ru/problem.aspx?space=1&num=1992 Description Yoda: Visit I will the c ...

  2. URAL 1992 CVS 可持久化链栈

    http://www.cnblogs.com/tedzhao/archive/2008/11/12/1332112.html 看这篇的链表部分的介绍应该就能理解“可持久化”了 动态分配内存的会T,只能 ...

  3. URAL 1992 CVS 链表

    不更改链表结构,只是添加节点,没有删除节点.通过记录和更改标记来模拟题意的插入和删除,复制 指针模拟链表: 预开指针,存在M[]中,可以提高效率 #include<functional> ...

  4. 后缀数组 POJ 3974 Palindrome && URAL 1297 Palindrome

    题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串 ...

  5. ural 2071. Juice Cocktails

    2071. Juice Cocktails Time limit: 1.0 secondMemory limit: 64 MB Once n Denchiks come to the bar and ...

  6. ural 2073. Log Files

    2073. Log Files Time limit: 1.0 secondMemory limit: 64 MB Nikolay has decided to become the best pro ...

  7. ural 2070. Interesting Numbers

    2070. Interesting Numbers Time limit: 2.0 secondMemory limit: 64 MB Nikolay and Asya investigate int ...

  8. ural 2069. Hard Rock

    2069. Hard Rock Time limit: 1.0 secondMemory limit: 64 MB Ilya is a frontman of the most famous rock ...

  9. ural 2068. Game of Nuts

    2068. Game of Nuts Time limit: 1.0 secondMemory limit: 64 MB The war for Westeros is still in proces ...

随机推荐

  1. 路由器开发板上的TTL线连接方法

    手头有个MTK双频路由器的开发板,做工良好,但让人蛋疼的是,TTL线没有标注TX/RX/GND/VCC,这个小细节的缺失给使用带来了巨大麻烦.   网上搜了半天也没找到相关电路图,只好遍历测试找到正确 ...

  2. 编写一段程序,从标准输入读取string对象的序列直到连续出现两个相同的单词或者所有单词都读完为止。使用while循环一次读取一个单词,当一个单词连续出现两次是使用break语句终止循环。输出连续重复出现的单词,或者输出一个消息说明没有人任何单词是重复出现的。

    // test14.cpp : 定义控制台应用程序的入口点. // #include "stdafx.h" #include<iostream> #include< ...

  3. [转载]GDI+中发生一般性错误

    注:第一次写博客,把自己遇到的问题和收集的资料记录在博客上.在开发.NET应用中,使用 System.Drawing.Image.Save 方法而导致“GDI+ 中发生一般性错误”的发生,通常有以下三 ...

  4. 01-07-01【Nhibernate (版本3.3.1.4000) 出入江湖】并发控制

    Nhibernate 并发控制 [1]悲观并发控制 正在使用数据的操作,加上锁,使用完后解锁释放资源. 使用场景:数据竞争激烈,锁的成本低于回滚事务的成本 缺点:阻塞,可能死锁 [2]乐观并发控制: ...

  5. 01-04-01【Nhibernate (版本3.3.1.4000) 出入江湖】原生的SQL查询

    Nhibernate 支持原生的SQL查询 : /// <summary> /// 使用原生的SQL查询 /// </summary> /// <param name=& ...

  6. 一个Java程序员应该掌握的10项技能

    1.语法:必须比较熟悉,在写代码的时候IDE的编辑器对某一行报错应该能够根据报错信息知道是什么样的语法错误并且知道任何修正. 2.命令:必须熟悉JDK带的一些常用命令及其常用选项,命令至少需要熟悉:a ...

  7. hdoj 1596 find the safest road

    题目传送:http://acm.hdu.edu.cn/showproblem.php?pid=1596 分析:Dijkstra变体,最短路径判断计算方式:Safe(P) = s(e1)*s(e2)…* ...

  8. hdu1116

    http://acm.hdu.edu.cn/showproblem.php?pid=1116 #include<stdio.h> #include<math.h> #inclu ...

  9. Project Euler 83:Path sum: four ways 路径和:4个方向

    Path sum: four ways NOTE: This problem is a significantly more challenging version of Problem 81. In ...

  10. *[hackerrank]Cut the tree

    https://www.hackerrank.com/contests/w2/challenges/cut-the-tree 树分成两部分,求两部分差最小.一开始想多了,后来其实求一下总和,求一下分部 ...