Array analog linked list

4273. List merge

Given two single chain tables L1=a1→a2→…→an−1→an and L2=b1→b2→…→bm−1→bm.

If n≥2m, Your task is to reverse the shorter list , Then merge it into a longer linked list , Get the shape of a1→a2→bm→a3→a4→bm−1… Result .

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

Input

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 105 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 .

1≤N≤105

Output

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

The 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

AC Code

#include<bits/stdc++.h>

using namespace std;


const int N = 100010;
int e[N],ne[N];
int main(){
    int h1,h2,n;
    cin >> h1 >> h2 >> n;
    
    while(n--){
        int add, data, next;
        cin >> add >> data >> next;
        e[add] = data;
        ne[add] = next;
    }
    vector<pair<int , int >> a, b;
    for(int i = h1; i != -1 ; i = ne[i])
    a.push_back(make_pair(i,e[i]));
    
    for(int i = h2; i != -1; i = ne[i])
    b.push_back(make_pair(i,e[i]));
    
    
    
    vector<pair<int ,int>> c;
    if(a.size() < b.size()) swap(a,b);
    for(int i = 0 , j = b.size() - 1; i < a.size(); i += 2, j--){
        c.push_back(a[i]);
        if(i + 1 < a.size()) c.push_back(a[i + 1]);
        if(j >= 0) c.push_back(b[j]);
    }
    
    for(int i = 0; i < c.size(); i++){
         printf("%05d %d ",c[i].first,c[i].second);
         if(i + 1 < c.size())
         printf("%.5d\n",c[i + 1].first);
         else
         puts("-1");
    }
    return 0;
   
}