PAT A1074 Reversing Linked List

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 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

首先要读清楚题意 每k个就要反转 可能需要反转很多次

使用vector进行反转很便捷

贴ac代码

#include
#include 
#include
using namespace std;
const int N=100010;
int v[N],ne[N];
vectorq;
int main()
{
    int head,n,k;
    cin>>head>>n>>k;
    for(int i=0;i>a>>b>>c;
        v[a]=b,ne[a]=c;
    }
    for(int i=head;i!=-1;i=ne[i]) q.push_back(i);
    for(int i=0;i+k-1