Mergeable Stack


Time Limit: 2 Seconds      Memory Limit: 65536 KB

Given initially empty stacks, there are three types of operations:

  • 1 s v: Push the value onto the top of the -th stack.

  • 2 s: Pop the topmost value out of the -th stack, and print that value. If the -th stack is empty, pop nothing and print "EMPTY" (without quotes) instead.

  • 3 s t: Move every element in the -th stack onto the top of the -th stack in order.

    Precisely speaking, denote the original size of the -th stack by , and the original size of the -th stack by . Denote the original elements in the -th stack from bottom to top by , and the original elements in the -th stack from bottom to top by .

    After this operation, the -th stack is emptied, and the elements in the -th stack from bottom to top becomes . Of course, if , this operation actually does nothing.

There are operations in total. Please finish these operations in the input order and print the answer for every operation of the second type.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains two integers and ( ), indicating the number of stacks and the number of operations.

The first integer of the following lines will be ( ), indicating the type of operation.

  • If , two integers and ( , ) follow, indicating an operation of the first type.
  • If , one integer ( ) follows, indicating an operation of the second type.
  • If , two integers and ( , ) follow, indicating an operation of the third type.

It's guaranteed that neither the sum of nor the sum of over all test cases will exceed .

Output

For each operation of the second type output one line, indicating the answer.

Sample Input

2
2 15
1 1 10
1 1 11
1 2 12
1 2 13
3 1 2
1 2 14
2 1
2 1
2 1
2 1
2 1
3 2 1
2 2
2 2
2 2
3 7
3 1 2
3 1 3
3 2 1
2 1
2 2
2 3
2 3

Sample Output

13
12
11
10
EMPTY
14
EMPTY
EMPTY
EMPTY
EMPTY
EMPTY
EMPTY

Author: WENG, Caizhi
Source: The 18th Zhejiang University Programming Contest Sponsored by TuSimple

分析:

/*
有n个栈,q次操作
1 s t:将t压入第s个栈
2 s:第s个栈pop一个元素并打印
3 s t:栈t从底到顶压入s栈,并将t栈清空
注意用list模拟栈的操作,特别是栈的合并操作,采用的是splice函数,学习了!!!
*/
 
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[]= {,,,,,,,,,,,,};
int mon2[]= {,,,,,,,,,,,,};
int dir[][]={{,},{,-},{,},{-,}}; list<int> li[];
int main()
{
int t,n,q,op;
int index1,index2,v;
cin>>t;
while(t--)
{
scanf("%d %d",&n,&q);
for(int i=;i<=n;i++)
li[i].clear();
while(q--)
{
scanf("%d",&op);
if(op==)
{
scanf("%d %d",&index1,&v);
li[index1].push_back(v);
}else if(op==)
{
scanf("%d",&index1);
if(li[index1].empty())
{
printf("EMPTY\n");
}
else
{
printf("%d\n",li[index1].back());
li[index1].pop_back();
}
}else if(op==)
{
scanf("%d %d",&index1,&index2);
li[index1].splice(li[index1].end(),li[index2]);
}
}
}
return ;
}
/*
有n个栈,q次操作
1 s t:将t压入第s个栈
2 s:第s个栈pop一个元素并打印
3 s t:栈t从底到顶压入s栈,并将t栈清空 注意用list模拟栈的操作,特别是栈的合并操作,采用的是splice函数,学习了!!!
*/

ZOJ 4016 Mergeable Stack(利用list模拟多个栈的合并,STL的应用,splice函数!!!)的更多相关文章

  1. ZOJ 4016 Mergeable Stack(栈的数组实现)

    Mergeable Stack Time Limit: 2 Seconds      Memory Limit: 65536 KB Given  initially empty stacks, the ...

  2. ZOJ 4016 Mergeable Stack 链表

    Mergeable Stack Time Limit: 2 Seconds      Memory Limit: 65536 KB Given  initially empty stacks, the ...

  3. ZOJ - 4016 Mergeable Stack 【LIST】

    题目链接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4016 题意 模拟栈的三种操作 第一种 push 将指定元素压入指 ...

  4. ZOJ - 4016 Mergeable Stack (STL 双向链表)

    [传送门]http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4016 [题目大意]初始有n个空栈,现在有如下三种操作: (1) ...

  5. ZOJ 4016 Mergeable Stack(from The 18th Zhejiang University Programming Contest Sponsored by TuSimple)

    模拟题,用链表来进行模拟 # include <stdio.h> # include <stdlib.h> typedef struct node { int num; str ...

  6. Codeforces 1131 F. Asya And Kittens-双向链表(模拟或者STL list)+并查集(或者STL list的splice()函数)-对不起,我太菜了。。。 (Codeforces Round #541 (Div. 2))

    F. Asya And Kittens time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  7. [ZOJ 4016] Mergable Stack

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4016 直接用栈爆内存,看网上大神用数组实现的,构思巧妙,学习了! ...

  8. 利用正则表达式模拟计算器进行字符串的计算实现eval()内置函数功能

    代码感觉有点绕,刚开始学习python,相关知识点还没全部学习到,还请各位大神多多指教 import re # 定义乘法 def mul(string): mul1 = re.search('-?\d ...

  9. Mergeable Stack ZOJ - 4016(list)

    ZOJ - 4016 vector又T又M list是以链表的方式存储的 是一个双向链表 元素转移操作中,不能一个个遍历加入到s中,list独有的splic函数可以在常数时间内实现合并(并删除源lis ...

随机推荐

  1. HDU3359(SummerTrainingDay05-I 高斯消元)

    Kind of a Blur Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)To ...

  2. mysql小试题3

    查询结果:

  3. 【代码笔记】iOS-JQIndicatorViewDemo

    一,效果图. 二,工程图. 三,代码. #import "ViewController.h" #import "JQIndicatorView.h" @inte ...

  4. oop(Object Oriented Programming)

    嗯,昨天忙了一天没来及发,过年啊,打扫啊,什么搽窗户啊,拖地啊,整理柜子啊,什么乱七八糟的都有,就是一个字,忙. 好了,废话也不多说,把自己学到的放上来吧.嗯,说什么好呢,就说原型链啊 原型对象 每个 ...

  5. 数据库小组与UI小组第一次对接

    时间:2018.6.1,21:30 ~ 23:00 人员:除黄志鹏外全体成员,因为黄志鹏临时有事 工作内容: 主要为数据库小组与UI第二组对接,并将成果汇总到github仓库.另外UI第一组重构了代码 ...

  6. android:screenOrientation属性

    今天工作中发现一个activity的android:screenOrientation属性设置为behind,平时经常看到的是landscape.portrait,一时没有反应过来,故查了一下andr ...

  7. Windows2003系统如何设置能让两个人共用一个桌面同时远程控制?

    在windows 2003上,可以两人同时同一桌面控制一台服务器,交流非常方便. 解决方案: 两人都用终端远程登陆到服务器上,其中一人在“开始”--“管理工具”--“终端服务管理器”,选中对方的用户名 ...

  8. Azure 元数据服务:适用于 Windows VM 的计划事件(预览)

    计划事件是 Azure 元数据服务中的其中一个子服务. 它负责显示有关即将发生的事件(例如,重新启动)的信息,使应用程序可以为其做准备并限制中断. 它可用于所有 Azure 虚拟机类型(包括 PaaS ...

  9. python基础学习21----进程

    python中的多线程其实并不是真正的多线程,如果想要充分地使用多核CPU的资源,在python中大部分情况需要使用多进程. 进程与线程的使用有很多相似之处,有关线程方面的知识请参考https://w ...

  10. Excel思考问题的方式

    Excel思考问题的方式 一.写需求,说我要什么数据 好比如,现在咱们需要将第一周.第二周.第三周.第四周.….等E:E列里的"每一周的 第二个数值"提取出来.那么我们手动提取了几 ...