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 (≤) which is the total number of nodes, and a positive K (≤) 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<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

#define MAXSIZE 1000010

struct node
{
    int data;
    int next;
} node[MAXSIZE];

int List[MAXSIZE];

int main()
{
    int begin,n,k;
    cin >> begin >> n >> k;
    int Address,Data,Next;
    for (int i = ; i < n; i++)
    {
        cin >> Address >> Data >> Next;
        node[Address].data = Data;
        node[Address].next = Next;
    }

    int count = ;
    int p = begin;
    while (p != -)
    {
        List[count++] = p;
        p = node[p].next;
    }

    for(int i = ; i + k <= count; i += k)
    {
        reverse(&List[i],&List[i+k]);
    }
/*
    int i = 0;
    while (i + k <= count)
    {
        reverse(&List[i],&List[i+k]);
        i += k;
    }
*/
    for(int i = ; i < count; i++)
    {
        if(i < count - )
        {
            printf("%05d %d %05d\n",List[i],node[List[i]].data,List[i+]);
        }
        else
        {
            printf("%05d %d -1",List[i],node[List[i]].data);
        }
    }

    return ;
}