个人学习记录，代码难免不尽人意。

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

#include <cstdio>
#include<algorithm>
using namespace std;
struct Node{int address;int next;int data;bool flag;
}node[100010];
bool cmp(Node a,Node b){if(a.flag==false||b.flag==false){return a.flag>b.flag;}else{return a.data<b.data;}}
int main(){for(int i=0;i<100010;i++){node[i].flag=false;}int n,begin;scanf("%d %d",&n,&begin);for(int i=0;i<n;i++){int address,next;int data;scanf("%d %d %d",&address,&data,&next);node[address].address=address;node[address].data=data;node[address].next=next;//node[address].flag=true;这个地方不能直接赋true，因为题目中给的数据可能不在链表上！ } int count=0;int p=begin;while(p!=-1){node[p].flag=true;count++;p=node[p].next;} if(count==0)printf("0 -1");else{sort(node,node+100010,cmp);printf("%d %05d\n",count,node[0].address);for(int i=0;i<count;i++){if(i!=count-1){printf("%05d %d %05d\n",node[i].address,node[i].data,node[i+1].address);}else{printf("%05d %d -1\n",node[i].address,node[i].data);}}}
}

本题采用了静态链表的方法，需要注意的是在链表节点内部存储了其地址，因为链表排序只改变节点的next值，而其本身的address不发生改变。所以最后输出的时候，next输出的是物理存储上的下一个节点的address值而不是next值，需要习惯这种思维方式。