HDU  5818

Problem Description
A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the stack. The last entry which is inserted is the first one that will be removed. In another word, the operations perform in a Last-In-First-Out (LIFO) manner.
A mergeable stack is a stack with "merge" operation. There are three kinds of operation as follows:

- push A x: insert x into stack A
- pop A: remove the top element of stack A
- merge A B: merge stack A and B

After an operation "merge A B", stack A will obtain all elements that A and B contained before, and B will become empty. The elements in the new stack are rearranged according to the time when they were pushed, just like repeating their "push" operations in one stack. See the sample input/output for further explanation.
Given two mergeable stacks A and B, implement operations mentioned above.

 
Input
There are multiple test cases. For each case, the first line contains an integer N(0<N≤105), indicating the number of operations. The next N lines, each contain an instruction "push", "pop" or "merge". The elements of stacks are 32-bit integers. Both A and B are empty initially, and it is guaranteed that "pop" operation would not be performed to an empty stack. N = 0 indicates the end of input.
 
Output
For each case, print a line "Case #t:", where t is the case number (starting from 1). For each "pop" operation, output the element that is popped, in a single line.
 
Sample Input
4
push A 1
push A 2
pop A
pop A
9
push A 0
push A 1
push B 3
pop A
push A 2
merge A B
pop A
pop A
pop A
9
push A 0
push A 1
push B 3
pop A
push A 2
merge B A
pop B
pop B
pop B
0
 
Sample Output
Case #1:
2
1
Case #2:
1
2
3
0
Case #3:
1
2
3
0
 
思路:用C栈保存A、B栈的合并,可以再用一个D栈保存A和B的合并,然后再导到C栈;
 
代码如下:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
using namespace std;
struct Node
{
int x;
int t;
};
stack<Node*>A,B,C,D; int main()
{
int N,Case=;
char s[];
char x;
int c;
while(scanf("%d",&N)&&N)
{
printf("Case #%d:\n",Case++);
for(int i=;i<=N;i++)
{
scanf("%s",s);
if(s[]=='u')
{
scanf(" %c",&x);
scanf("%d",&c);
Node* a=new Node();
a->x=c;
a->t=i;
if(x=='A')
A.push(a);
else
B.push(a);
}
else if(s[]=='o')
{
scanf(" %c",&x);
if(x=='A')
{
if(A.empty())
{
printf("%d\n",C.top()->x);
C.pop();
}
else
{
printf("%d\n",A.top()->x);
A.pop();
}
}
else
{
if(B.empty())
{
printf("%d\n",C.top()->x);
C.pop();
}
else
{
printf("%d\n",B.top()->x);
B.pop();
}
}
}
else if(s[]=='e')
{
scanf(" %c",&x);
scanf(" %c",&x);
while(!A.empty()&&!B.empty())
{
if(A.top()->t>B.top()->t)
{
D.push(A.top());
A.pop();
}
else
{
D.push(B.top());
B.pop();
}
}
while(!A.empty())
{
D.push(A.top());
A.pop();
}
while(!B.empty())
{
D.push(B.top());
B.pop();
}
while(!D.empty())
{
C.push(D.top());
D.pop();
}
}
}
}
return ;
}

2016暑假多校联合---Joint Stacks (STL)的更多相关文章

  1. 2016暑假多校联合---Rikka with Sequence (线段树)

    2016暑假多校联合---Rikka with Sequence (线段树) Problem Description As we know, Rikka is poor at math. Yuta i ...

  2. 2016暑假多校联合---Windows 10

    2016暑假多校联合---Windows 10(HDU:5802) Problem Description Long long ago, there was an old monk living on ...

  3. 2016暑假多校联合---Substring(后缀数组)

    2016暑假多校联合---Substring Problem Description ?? is practicing his program skill, and now he is given a ...

  4. 2016暑假多校联合---To My Girlfriend

    2016暑假多校联合---To My Girlfriend Problem Description Dear Guo I never forget the moment I met with you. ...

  5. 2016暑假多校联合---A Simple Chess

    2016暑假多校联合---A Simple Chess   Problem Description There is a n×m board, a chess want to go to the po ...

  6. 2016暑假多校联合---Another Meaning

    2016暑假多校联合---Another Meaning Problem Description As is known to all, in many cases, a word has two m ...

  7. 2016暑假多校联合---Death Sequence(递推、前向星)

    原题链接 Problem Description You may heard of the Joseph Problem, the story comes from a Jewish historia ...

  8. 2016暑假多校联合---Counting Intersections

    原题链接 Problem Description Given some segments which are paralleled to the coordinate axis. You need t ...

  9. 2016暑假多校联合---GCD

    Problem Description Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). ...

随机推荐

  1. Java线程:线程的交互

      一.线程交互的基础知识   SCJP所要求的线程交互知识点需要从java.lang.Object的类的三个方法来学习:    void notify()           唤醒在此对象监视器上等 ...

  2. [常见问题]解决创建servlet 找不到webservlet包.

    今天在创建一个springmvc项目的时候发现 使用的HttpServletRequest不起作用, 提示需要映入 jar文件, 于是便有了今天的这个问题: 百度了下才发现 项目需要导入Runtime ...

  3. VS2012 easyui datagrid url访问之坑

    VS2012 easyui datagrid url访问之坑 url属性放的是地址的话 返回的json格式必须有 total 和 rows,如下: {"total":2," ...

  4. git回滚到任意版本

    git回滚到任意版本 先显示提交的log $ git log -3 commit 4dc08bb8996a6ee02f Author: Mark <xxx@xx.com> Date: We ...

  5. Java时间日期格式转换

    1.这个是系统自动默认的时间格式,或者说是美式格式: Long time = System.currentTimeMillis();                Date date = new Da ...

  6. POJ1014 解题报告(DFS)

    题目在此:http://poj.org/problem?id=1014 要看清题意呢,题中要求输入的是价值分别为1,2,3,4,5,6的大理石的个数,而不是6块价值为输入数字的大理石!选这个题主要想练 ...

  7. 真正的mybatis_redis二级缓存

    网上流传的代码缓存失效存在严重问题. 思路....以后再细说 目前的方案还不够完美,失效力度控制不够细. 主要代码 import java.io.BufferedReader; import java ...

  8. KlayGE 4.4中渲染的改进(四):SSSSS

    转载请注明出处为KlayGE游戏引擎,本文的永久链接为http://www.klayge.org/?p=2774 本系列的上一篇提到了KlayGE 4.4将会出现的高质量地形渲染.本篇仍讲一个新功能, ...

  9. gridview里日期显示格式

    Text='<%#Bind("EndDate","{0:yyyy-MM-dd}") %>'

  10. java中final注意的问题

    public class Test{ public static void main(String[] args){ Person p = new Person(); } } /* 4.修饰的变量是一 ...