List merging (summer vacation daily question 3)

Given two single chain tables $L_1=a_1\to a_2\to \dots \to a_{n-1}\to a_n$ and
$L_2=b_1\to b_2\to \dots \to b_{m-1}\to b_m$.

If $n\ge 2m$, Your task is to reverse the shorter list , Then merge it into a longer linked list , Get the shape of $a_1\to a_2\to b_m\to a_3\to a_4\to b_{m-1}\dots$ Result .

For example, give two linked lists as $6\to 7$ and $1\to 2\to 3\to 4\to 5$, You should output $1\to 2\to 7\to 3\to 4\to 6\to 5$.

Add
This question may contain nodes that are not in the two single linked lists , These nodes need not be considered .

Input format
Input first gives two linked lists in the first row $L1$ and $L2$ The address of the header node of , And positive integers $N$, That is, the total number of nodes given .

The address of a node is $5$ Nonnegative integer of digits （ May contain leading $0$）, Empty address NULL use −1 Express .

And then $N$ That's ok , Each line gives the information of a node in the following format ：

Address Data Next

among Address Is the address of the node ,Data No more than $10^5$ The positive integer ,Next Is the address of the next node .

The title ensures that there is no empty linked list , And the longer list is at least twice as long as the shorter list .

Output format
Output the result linked list in order , Each node occupies a row , The format is the same as the input .

Data range
$1\le N\le 10^5$

sample input ：

00100 01000 7
02233 2 34891
00100 6 00001
34891 3 10086
01000 1 02233
00033 5 -1
10086 4 00033
00001 7 -1

sample output ：

01000 1 02233
02233 2 00001
00001 7 34891
34891 3 10086
10086 4 00100
00100 6 00033
00033 5 -1

---

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

const int N = 1000010;

int n, h1, h2;
int e[N], ne[N];

int main(){
    
    
    cin >> h1 >> h2 >> n;
    
    int address, data, neaddress;
    for(int i = 0; i < n; i++){
    
        cin >> address >> data >> neaddress;
        e[address] = data;
        ne[address] = neaddress;
    }
    
    vector<int> v1, v2, v3;
    for(int i = h1; ~i; i = ne[i])
        v1.push_back(i);
    for(int i = h2; ~i; i = ne[i])
        v2.push_back(i);
    
    if(v1.size() < v2.size()) swap(v1, v2);
    
    reverse(v2.begin(), v2.end());
    for(int i = 0; i < v1.size(); i++){
    
        v3.push_back(v1[i]);
        if(i % 2 && i / 2 < v2.size()) v3.push_back(v2[i/2]);
    }
    
    for(int i = 0; i < v3.size(); i++){
    
        
        printf("%05d %d ", v3[i], e[v3[i]]);
        if(i == v3.size() - 1) puts("-1");
        else printf("%05d\n", v3[i+1]);
    }
    
}