02-线性结构3 Reversing Linked List (25分)

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 $N$ (\le 10^5≤10​5​​) which is the total number of nodes, and a positive $K$ ($\le N$) 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 Address is the position of the node, Data is 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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

注意：测试的数据中可能有不在链表上的结点，此结点应直接删除，不予出现，所以数据输入之后，需要建立一个数组，将所有链接起来的结点连在一起，不在链表上的结点直接跳过，之后再进行下一步操作。

#include<cstdio>#include<algorithm>using namespace std;#define maxsize 1000010struct node                                                     // 定义一个结构体数组{int data;int next;}node[maxsize];int List[maxsize];                        int main(){int Adr,Data,Next;int first,N,K,i;scanf("%d%d%d",&first,&N,&K);for( i=0;i<N;i++)                                            //输入数据{scanf("%d%d%d",&Adr,&Data,&Next);node[Adr].data=Data;node[Adr].next=Next;}int p=first;int j=0;while(p!=-1)                                                  //将所有在链表上的结点存放在一个数组中，{List[j++]=p;p=node[p].next;}i=0;while(i+K<=j)                                                 // 每k个进行反转并循环{reverse(&List[i],&List[i+K]);i=i+K;}for(i=0;i<j-1;i++){printf("%05d %d %05d\n",List[i],node[List[i]].data,List[i+1]);   //  按格式输出}printf("%05d %d -1\n",List[i],node[List[i]].data);return 0;}