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

here Address is the position of the node, Data is an integer in [−105 ,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

cpp代码

#include<iostream>
#include<unordered_map>
#include<vector>
using namespace std;
unordered_map<int,int> date,ne;
vector<pair<int,int>> neg,pre,now;
vector<pair<int,int>> node,res;
int main(){
    int beg,n,m;
    cin>>beg>>n>>m;
    for(int i=0;i<n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        date[a]=b;
        ne[a]=c;
    }
    for(int i=beg;~i;i=ne[i]){
       node.push_back({i,date[i]});
    }
    for(auto i:node){
        int x=i.first,y=i.second;
        if(y<0)neg.push_back({x,y});
        else if(y>=0&&y<=m)pre.push_back({x,y});
        else now.push_back({x,y});
    }
    for(auto i:neg)
       res.push_back({i.first,i.second});
    for(auto i: pre)
       res.push_back({i.first,i.second});
    for(auto i: now)
        res.push_back({i.first,i.second});
   
    for(int i=0;i<res.size();i++){
        if(i!=res.size()-1)printf("%05d %d %05d\n",res[i].first,res[i].second,res[i+1].first);
        else printf("%05d %d -1\n",res[i].first,res[i].second);
    }
    return 0;
}