給定一個帶整數鍵值的單鏈表L，本題要求你編寫程式，刪除那些鍵值的絕對值有重複的結點。即對任意鍵值K，只有鍵值或其絕對值等於K的第一個結點可以被保留。同時，所有被刪除的結點必須被儲存在另外一個連結串列中。例如：另L為21→-15→-15→-7→15，則你必須輸出去重後的連結串列21→-15→-7、以及被刪除的連結串列-15→15。

輸入格式：

輸入第一行包含連結串列第一個結點的地址、以及結點個數N（<= 10⁵ 的正整數）。結點地址是一個非負的5位整數，NULL指標用-1表示。

隨後N行，每行按下列格式給出一個結點的資訊：

Address Key Next

其中Address是結點的地址，Key

輸出格式：

首先輸出去重後的連結串列，然後輸出被刪除結點組成的連結串列。每個結點佔一行，按輸入的格式輸出。

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

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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
 int vis[1001100];
    int mapp[1001100],mappp[1001100];
    int keep1[1001100],keep2[1001100];
int main()
{
    int a,b,c;
    int vv1=0,vv2=0;
    memset(vis,0,sizeof(vis));
    int m,n;
    scanf("%d%d",&m,&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        mapp[a]=b;
        mappp[a]=c;
    }
    int pp;
    pp=m;
   while(mappp[pp]!=-1)
   {
       int flag=0;
       if(mapp[pp]<0)
       {
           flag=1;
           mapp[pp]=-mapp[pp];
       }
       if(vis[mapp[pp]]==0)
       {
           if(vv2>0)
            keep2[vv2-1]=mappp[pp];
           if(flag==1)
           {
           keep1[vv1++]=pp;
           keep1[vv1++]=-mapp[pp];
           keep1[vv1++]=mappp[pp];
           vis[mapp[pp]]=1;
           pp=mappp[pp];
           }
          else
          {
           keep1[vv1++]=pp;
           keep1[vv1++]=mapp[pp];
           keep1[vv1++]=mappp[pp];
           vis[mapp[pp]]=1;
           pp=mappp[pp];
          }
       }
       else
       {
           keep1[vv1-1]=mappp[pp];
           if(flag==1)
           {
           keep2[vv2++]=pp;
           keep2[vv2++]=-mapp[pp];
           keep2[vv2++]=mappp[pp];
           pp=mappp[pp];
           }
           else
           {
           keep2[vv2++]=pp;
           keep2[vv2++]=mapp[pp];
           keep2[vv2++]=mappp[pp];
           pp=mappp[pp];
           }
       }
   }
   int flag1=0;
   if(mapp[pp]<0)
   {
       flag1=1;
       mapp[pp]=-mapp[pp];
   }
   if(vis[mapp[pp]]==1)
   {
       if(flag1==0)
       {
           keep2[vv2++]=pp;
           keep2[vv2++]=mapp[pp];
           keep2[vv2++]=mappp[pp];
           keep1[vv1-1]=-1;
       }
       else
       {
           keep2[vv2++]=pp;
           keep2[vv2++]=-mapp[pp];
           keep2[vv2++]=mappp[pp];
           keep1[vv1-1]=-1;
       }
   }
   else
   {
       if(!flag1)
       {
           keep1[vv1++]=pp;
           keep1[vv1++]=mapp[pp];
           keep1[vv1++]=mappp[pp];
           keep2[vv2-1]=-1;
       }
       else
       {
           keep1[vv1++]=pp;
           keep1[vv1++]=-mapp[pp];
           keep1[vv1++]=mappp[pp];
           keep2[vv2-1]=-1;
       }
   }
   for(int i=1;i<=vv1;i++)
   {
       if(keep1[i-1]==-1)
        printf("-1");
       else if(i%3==1||i%3==0)
        printf("%05d",keep1[i-1]);
         else if(i%3==2)
        printf("%d",keep1[i-1]);
        if(i%3)
            printf(" ");
        else
            printf("\n");
   }
   for(int i=1;i<=vv2;i++)
   {
       if(keep2[i-1]==-1)
        printf("-1");
       else if(i%3==1||i%3==0)
        printf("%05d",keep2[i-1]);
       else if(i%3==2)
        printf("%d",keep2[i-1]);
        if(i%3)
            printf(" ");
        else
            printf("\n");
   }
    return 0;
}