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<iostream>
#include<unordered_map>
#include<vector>
using namespace std;
struct node{
	int address;
	int data;
};
int main(){
	unordered_map<int,int> m1;
	unordered_map<int,int> m2;
	vector<int> book;
	vector<node> ans;
	int h;
	int n,num;
	scanf("%d %d %d",&h,&n,&num);
	for(int i = 0;i < n;i++){
		int x,y,z;
		scanf("%d %d %d",&x,&y,&z);
		m1[x] = z;
		m2[x] = y;
	}
	int pos = h;
	for(;pos != -1;pos= m1[pos]){
		if(m2[pos] < 0){
			ans.push_back(node{pos,m2[pos]});
		}
	}
	pos = h;
	for(;pos != -1;pos= m1[pos]){
		if(m2[pos] >= 0 && m2[pos] <= num){
			ans.push_back(node{pos,m2[pos]});
		}
	}
	pos = h;
	for(;pos != -1;pos= m1[pos]){
		if(m2[pos] > num){
			ans.push_back(node{pos,m2[pos]});
		}
	}
	for(int i = 0;i < ans.size();i++){
		printf("%05d %d",ans[i].address,ans[i].data);
		if(i == ans.size()-1){
			printf(" -1\n");
		}else{
			printf(" %05d\n",ans[i+1].address);
		}
	}
	return 0;
}