题目来源：PTA02-线性结构2Reversing Linked List(25分)

　　Question:Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.

　　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 (≤N) which is the length of the sublist to be reversed. 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

　　where Addressis the position of the node, Datais an integer, and Nextis the position of the next node.

　　Output Specification:

　　For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

　　Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

　　Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

　　Show the code:

#include<iostream>
#include<algorithm>    //使用algorithm的reverse函数进行反转
using namespace std;

const int MaxSize = 100000+10;

struct LinkNode        //节点，因为给定了地址值，用数组下标来表示地址，用next来指向下一地址(而不是指针)
{
    int data;
    int next;

}LinkNode[MaxSize];

int SeqList[MaxSize];    //简单的顺序表，用来存放节点间的(地址)连接关系

int main()
{
    int FirstAddr,N,K;   //N是输入的节点数，但这些节点不一定都能连在一起
    cin>>FirstAddr>>N>>K;//输入第一行
    int Address,Data,Next,i;
    for(i=0;i<N;i++)
    {
        cin>>Address>>Data>>Next;
        LinkNode[Address].data = Data;
        LinkNode[Address].next = Next;
    }
    int List_n=0;       //把能连在一起的节点地址存入顺序表，List_n表示这些节点的数目
    int p=FirstAddr;    //p指示当前节点
    while(p!=-1)        
    {
        SeqList[List_n++] = p;
        p = LinkNode[p].next;
    }
    i=0;
    while(i+K<=List_n)
    {
        reverse(&SeqList[i],&SeqList[i+K]);//Reverses the order of the elements in the range [first,last)
        i=i+K;
    }
    for (i = 0;i < List_n-1;i++)    //输出逆转后的“链表”
    {
        printf("%05d %d %05d\n", SeqList[i], LinkNode[SeqList[i]].data, SeqList[i+1]);//%05d为输出5位整数，不足在前面补0
    }
    printf("%05d %d -1\n", SeqList[i], LinkNode[SeqList[i]].data);    //输出逆转后的“链表”的最后一项
    return 0;
}