它的测试数据有可能有分裂节点，所以需要计算实际所给链表的长度

#include<bits/stdc++.h>
using namespace std;
struct Node{
	int val;
	int next;
}x[100005];
int main(){
    int j=0;
	int start;
	int n;
	int ad1,num,ad2;
	cin>>start>>n;
	for(int i=0;i<n;i++){
		cin>>ad1>>num>>ad2;
		x[ad1].val=num;
		x[ad1].next=ad2;		
	}
    /*解决多余节点的代码*/
    int first=start;
    while(first!=-1)
    {
        first=x[first].next;
        j++;
    }
    n=j;
/*解决多余节点的代码*/
	int p=start;
	vector<int> pre;
	vector<int> post;
	int mid=n/2;
	int count=1;
//	cout<<x[0].val<<" "<<x[0].next<<endl;
//	cout<<x[100].val<<" "<<x[100].next<<endl;
	while(p!=-1){
		if(count<=mid){
			pre.push_back(p);
		}
		else{
			post.push_back(p);
		}
		p=x[p].next;
		count++;
	}
	
	
	int ori=mid;
	vector<int> res;
	if(n % 2==0){
		while(mid){
			res.push_back(post[mid-1]);
			res.push_back(pre[ori-mid]);
			mid--;
		}
	}else{
		while(mid){
			res.push_back(post[mid]);
			res.push_back(pre[ori-mid]);
			mid--;
		}
		res.push_back(post[0]);
	}
	for(int i=0;i<n-1;i++){
		printf("%05d %d %05d\n",res[i],x[res[i]].val,res[i+1]);
	}
	printf("%05d %d -1",res[n-1],x[res[n-1]].val);
	return 0;
}