题目地址
Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10
​5
​​ ) which is the total number of nodes, and a positive K (≤10
​3
​​ ). The address of a node is a 5-digit nonnegative integer, and NULL is represented by −1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer in [−10
​5
​​ ,10
​5
​​ ], and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:
For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:
00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

Sample Output:
33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

#include <iostream>
#include <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=100010,inf=100000000;
struct node{
    int data,add,next;
}s[maxn];
int n,k,head;
vector<node>ans1,ans2,ans3;
int main(){
    int add;
    cin>>head>>n>>k;
    for(int i=0;i<n;i++){
        cin>>add;cin>>s[add].data>>s[add].next;
        s[add].add=add;
    }
    for(int i=head;i!=-1;i=s[i].next){
        if(s[i].data<0) ans1.push_back(s[i]);
        else if(s[i].data<=k) ans2.push_back(s[i]);
        else ans3.push_back(s[i]);
    }
    for(int i=0;i<ans2.size();i++) ans1.push_back(ans2[i]);
    for(int i=0;i<ans3.size();i++) ans1.push_back(ans3[i]);
    for(int i=0;i<ans1.size();i++){
        printf("%05d %d ",ans1[i].add,ans1[i].data);
        if(i!=ans1.size()-1) printf("%05d\n",ans1[i+1].add);
        else cout<<"-1"<<endl;
    }
}