1133 Splitting A Linked List

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 (≤103). 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 [−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.

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 
#include 
using namespace std;
int n, k, t, addr, num, n1, n2, n3;
int Data[100010], Next[100010], List[100010], N1[100010], N2[100010], N3[100010];
 
int main() {
	cin >> addr >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> t;
		cin >> Data[t] >> Next[t];
	}
	t = addr;
	while (t != -1) {
		List[num++] = t;
		t = Next[t];
	}
	for (int i = 0; i < num; i++) {
		if (Data[List[i]] < 0) {
			N1[n1++] = List[i];
		} else if (Data[List[i]] >= 0 && Data[List[i]] <= k) {
			N2[n2++] = List[i];
		} else {
			N3[n3++] = List[i];
		}
	}
	num = 0;
	for (int i = 0; i < n1; i++) {
		List[num++] = N1[i];
	}
	for (int i = 0; i < n2; i++) {
		List[num++] = N2[i];
	}
	for (int i = 0; i < n3; i++) {
		List[num++] = N3[i];
	}
	for (int i = 0; i < num - 1; i++) {
		cout << setw(5) << setfill('0') << List[i] << ' ' << Data[List[i]] << ' ' << setw(5) << setfill('0') << List[i + 1] <<
		     endl;
	}
	cout << setw(5) << setfill('0') << List[num - 1] << ' ' << Data[List[num - 1]] << ' ' << -1;
	return 0;
}