1133 Splitting A Linked List (25分)

1133 Splitting A Linked List (25分)

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
using namespace std;
struct Node{
	int data,next;
}node[100002];
vector<int>temp,ans,temp1,temp2,temp3;
int main(){
	int first,N,K,address;
	scanf("%d%d%d",&first,&N,&K);
	for(int i=0;i<N;i++){
		scanf("%d",&address);
		scanf("%d%d",&node[address].data,&node[address].next);
	}
	while(first!=-1){
		temp.push_back(first);
		first=node[first].next;
	}
	for(int i=0;i<temp.size();i++){
		if(node[temp[i]].data<0)temp1.push_back(temp[i]);
		else if(node[temp[i]].data<=K)temp2.push_back(temp[i]);
		else temp3.push_back(temp[i]);
	}
	for(int i=0;i<temp1.size();i++){
		ans.push_back(temp1[i]);
	}
	for(int i=0;i<temp2.size();i++){
		ans.push_back(temp2[i]);
	}
	for(int i=0;i<temp3.size();i++){
		ans.push_back(temp3[i]);
	}
	for(int i=0;i<ans.size()-1;i++){
		printf("%05d %d %05d\n",ans[i],node[ans[i]].data,ans[i+1]);		
	}
	printf("%05d %d -1\n",ans[ans.size()-1],node[ans[ans.size()-1]].data);
	return 0;
}