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 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 (≤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 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;
}