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 (≤$105$) 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:

where Address is the position of the node, Data is an integer in [−$105$ ,$105$], 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.

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

## 解答

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<vector>
#include<unordered_map>

using namespace std;

struct node { int add, key, next; };

int main()
{
unordered_map<int, node> list;
scanf("%d %d %d", &head, &num, &K);
for (int i = 0; i < num; i++) {
scanf("%d %d %d", &add, &key, &next);
}
vector<node> v, ans;

for (int tail = head; tail != -1; tail = list[tail].next)
v.push_back(list[tail]);

for (int i = 0; i < v.size(); i++)
if (v[i].key < 0) ans.push_back(v[i]);
for (int i = 0; i < v.size(); i++)
if (v[i].key >= 0 && v[i].key <= K) ans.push_back(v[i]);
for (int i = 0; i < v.size(); i++)
if (v[i].key > K) ans.push_back(v[i]);

for (int i = 0; i < v.size() - 1; i++)