TIANTI race - l2-002 linked list weight removal (25 points)

L2-002 The chain watch is weightless (25 branch )

Given a linked list with integer keys  L, You need to delete the key node where the absolute value is repeated . For each key value  K, Only the first absolute value is equal to  K  The nodes of are reserved . meanwhile , All deleted nodes must be saved in another linked list . For example, given  L  by 21→-15→-15→-7→15, You need to output the list after de duplication 21→-15→-7, And the deleted list -15→15.

Input format ：

Enter... On the first line L The address of the first node and a positive integer  N（≤105, Is the total number of nodes ）. The address of a node is nonnegative 5 An integer , Empty address NULL use  −1  To express .

And then  N  That's ok , Each line describes a node in the following format ：

 Address   Key value   Next node 

among Address Is the address of the node , Key value It's absolutely no more than 104 The integer of , Next node Is the address of the next node .

Output format ：

First output the list after de duplication , Then output the deleted list . Each node occupies a row , Output... In input format .

sample input ：

00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854

sample output ：

00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=10100;

struct node// Store three values for each row 
{
    string idx;
    int e;
    string ne;
};

map<string,node>mp;// Actually store each row , Subscript is address 
int book[100010];// Subscript to be i The absolute value of appears several times 



int main()
{
    string s;
    int n;
    cin>>s>>n;
    
    for(int i=0;i<n;i++)
    {
        string a,c;
        int b;
     
        cin>>a>>b>>c;
        mp[a]={a,b,c};
          
    }
    
    
    
    vector<node>ans1,ans2;// Store the deleted linked list and the deleted linked list respectively 
    
 //   cout<<n<<endl;
   // ans1[0].idx=s;
    
    for(int i=1;i<=n;i++)// Deal with this n Data 
    {
    //	cout<<",,,,"<<endl;
        node t=mp[s];
        if(book[abs(t.e)]>=1)// There have been   Just store it in ans2 Inside  
		{
			if(ans2.size()>=1)
			{
				ans2[ans2.size()-1].ne=t.idx;
			}
			ans2.push_back(t); 
		}
		else// There has been no ans1 Middle and processed pointer  
		{
			book[abs(t.e)]++;
			if(ans1.size()>=1)
			{
				ans1[ans1.size()-1].ne=t.idx;
			}
			ans1.push_back(t);
		}
		if(t.ne=="-1")break;
		s=t.ne;
		
    }
    
    
    ans1[ans1.size() - 1].ne = "-1";
    if (ans2.size() != 0)
       ans2[ans2.size() - 1].ne = "-1";
    

    
    
    for(auto i:ans1)
    cout<<i.idx<<" "<<i.e<<" "<<i.ne<<endl;
    for(auto i:ans2)
    cout<<i.idx<<" "<<i.e<<" "<<i.ne<<endl;
    
    
    return 0;
}