02-線性結構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 Kelements 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 (≤10​5​​) 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 Address is the position of the node, Data is an integer, and Next is 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

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=1e6+15;
struct Node
{
    int next;
    int data;
 
}node[maxn+1];
 
int List[maxn+1];
int main()
{
 
    int first,n,k;
    int ddata,nnext,add;
    cin>>first>>n>>k;
    int p=first;
    for(int i=0; i<n; i++)
    {
       cin>>add>>ddata>>nnext;
       node[add].data=ddata;
       node[add].next=nnext;
    }
 
    int j=0;
 
    while(p!=-1)
    {
        List[j++]=p;
        p=node[p].next;
    }
    int i=0;
    while((i+k)<=j)
    {
        reverse(&List[i],&List[i+k]);
        i=i+k;
    }
    for(int i=0; i<j-1; i++)
        printf("%05d %d %05d\n",List[i],node[List[i]].data,List[i+1]);
    printf("%05d %d -1\n",List[j-1],node[List[j-1]].data);
    return 0;
}