ZOJ 4016 Mergeable Stack（利用list模拟多个栈的合并，STL的应用，splice函数！！！）

Mergeable Stack

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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

• 1 s v: Push the value -th stack.

• 2 s: Pop the topmost value out of the

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

  Precisely speaking, denote the original size of the .

  After this operation, the , 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 ), indicating the number of stacks and the number of operations.

The first integer of the following ), indicating the type of operation.

• If ) follow, indicating an operation of the first type.
• If ) follows, indicating an operation of the second type.
• If ) follow, indicating an operation of the third type.

It's guaranteed that neither the sum of .

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

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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[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

list<int> li[300005];
int main()
{
    int t,n,q,op;
    int index1,index2,v;
    cin>>t;
    while(t--)
    {
        scanf("%d %d",&n,&q);
        for(int i=0;i<=n;i++)
            li[i].clear();
        while(q--)
        {
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d %d",&index1,&v);
                li[index1].push_back(v);
            }else if(op==2)
            {
                scanf("%d",&index1);
                if(li[index1].empty())
                {
                    printf("EMPTY\n");
                }
                else
                {
                    printf("%d\n",li[index1].back());
                    li[index1].pop_back();
                }
            }else if(op==3)
            {
                scanf("%d %d",&index1,&index2);
                li[index1].splice(li[index1].end(),li[index2]);
            }
        }
    }
    return 0;
}
/*
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！！！
*/