ZOJ 4016 Mergeable Stack 链表

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  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

题意 t组样例 每组 n个栈 q个操作 操作1 将 val 放到第i个栈顶
　　　　　　　　　　　　　　　　　操作2 输出 第 i 个栈的栈顶 弹出栈顶 为空 输出EMPTY
　　　　　　　　　　　　　　　　　操作3 将 第 j个栈 整个栈 放到第i个栈的栈顶

解析 用栈的话会MLE TLE 要用到c++容器 list 操作三就是o(1)复杂度 内存也没有问题

AC代码

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <utility>
#include <stack>
#include <list>
using namespace std;
const int maxn = 3e5+,inf = 0x3f3f3f3f, mod = ;
const double epx = 1e-;
const double PI = acos(-1.0);
typedef long long ll;
list<int> l[maxn];
int main()
{
    int t,n,q;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&q);
        for(int i=;i<=n;i++)
            l[i].clear();
        while(q--)
        {
            int m,i,j,val;
            scanf("%d",&m);
            if(m==)
            {
                scanf("%d%d",&i,&val);
                l[i].push_back(val);
            }
            else if(m==)
            {
                scanf("%d",&i);
                if(l[i].empty())
                    printf("EMPTY\n");
                else
                {
                    printf("%d\n",l[i].back());
                    l[i].pop_back();
                }
            }
            else
            {
                scanf("%d%d",&i,&j);
                l[i].splice(l[i].end(),l[j]);
            }
        }
    }
}

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