A1052 Linked List Sorting 链表的题坑太深

【题目描述】
1052 Linked List Sorting
A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.
【Input Specification】
Each input file contains one test case. For each case, the first line contains a positive N (<10​5​​) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by −1.
Then N lines follow, each describes a node in the format:
Address Key Next
where Address is the address of the node in memory, Key is an integer in [−10​5​​,10​5​​], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.
【Output Specification】
For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.
【Sample Input】
5 00001
11111 100 -1
00001 0 22222
33333 100000 11111
12345 -1 33333
22222 1000 12345
【Sample Output】
5 12345
12345 -1 00001
00001 0 11111
11111 100 22222
22222 1000 33333
33333 100000 -1

【坑】
链表的题都不难，但是不了解的人很难做全对，几乎在题库中能找到的甲级乙级的关于链表的题，都涉及到一个问题：无效节点，如果想不到这个的话很难AC，我就是这样，哭唧唧！（下面的代码是可以AC的代码）
【code】

#include<iostream>
#include<algorithm>
using namespace std;
struct node{
    int address,data,next,flag=0;
};
typedef struct node NODE;
NODE jd[100005],rs[100005];
bool cmp(NODE x,NODE y){
    return x.data<y.data;
}
int hashtable[100005]={0};
int main(){
    int N,i,start,cnt=0;
    scanf("%d%d",&N,&start);
    for(i=0;i<N;i++){
        scanf("%d%d%d",&jd[i].address,&jd[i].data,&jd[i].next);
        hashtable[jd[i].address]=i;
    }
    for(i=0;i<N&&start!=-1;i++){
        jd[hashtable[start]].flag=1;
        rs[i]=jd[hashtable[start]];
        start=jd[hashtable[start]].next;
        cnt++;
    }
    sort(rs,rs+cnt,cmp);
    if(cnt==0){
        printf("0 -1");
        return 0;
    }
    else{
        printf("%d %05d\n",cnt,rs[0].address);
        for(i=0;i<cnt-1;i++)
            printf("%05d %d %05d\n",rs[i].address,rs[i].data,rs[i+1].address);
        printf("%05d %d -1\n",rs[i].address,rs[i].data);
    }
    return 0;
}