HDU5818 Joint Stacks_acm mergeable stack-CSDN博客

本文链接：https://blog.csdn.net/Jelly_acm/article/details/75094333

本文介绍了一种特殊的数据结构——可合并栈，并详细解释了如何通过编程实现其基本操作，包括push、pop及merge。在merge操作中，两个栈的内容会根据元素的推入顺序进行合并。文章提供了一个具体的实现示例，帮助读者理解这些操作的具体流程。

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Joint Stacks

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1815 Accepted Submission(s): 815

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

Author

SYSU

Source

2016 Multi-University Training Contest 7

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题目的意思给出两个栈A B(初始时为空)，有三种操作：
push、pop、merge.
其中merge是按照A B中元素进栈的相对顺序来重排的.

思路：因为题目说了不会对空栈弹出，所以开三个优队（栈也可以）前两个分别存A和B，第三个公用，插入删除还是对AB操作，合并把AB情况放到C中，每次输入时先判AB是否是空，不为空则输出AB顶，否则这个元素肯定在C中输出即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,x;
char s1[10],s2[5],s3[5];
int cnt [1000005],fl[1000006];
struct node
{
    int t,val;
    friend bool operator <(const node &a,const node &b)
    {
        return a.t<b.t;
    }
} pre;

int main()
{
    int cas=0;
    while(~scanf("%d",&n)&&n)
    {
        printf("Case #%d:\n",++cas);
        priority_queue<node>q1,q2,q3;
        for(int i=1; i<=n; i++)
        {
            scanf("%s",s1);
            if(!strcmp(s1,"push"))
            {
                scanf("%s%d",s2,&x);
                pre.t=i,pre.val=x;
                if(!strcmp(s2,"A")) q1.push(pre);
                else q2.push(pre);
            }
            else if(!strcmp(s1,"pop"))
            {
                scanf("%s",s2);
                if(!strcmp(s2,"A"))
                {
                    if(!q1.empty())
                        printf("%d\n",q1.top().val),q1.pop();
                    else
                        printf("%d\n",q3.top().val),q3.pop();
                }
                else
                {
                    if(!q2.empty())
                    printf("%d\n",q2.top().val),q2.pop();
                     else
                        printf("%d\n",q3.top().val),q3.pop();
                }
            }
            else
            {
                scanf("%s%s",s2,s3);
                    while(!q2.empty())
                        q3.push(q2.top()),q2.pop();
                    while(!q1.empty())
                        q3.push(q1.top()),q1.pop();
            }
        }
    }
    return 0;
}