题目地址

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 (≤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 <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=100020,inf=100000000;
struct node{
    int address,data,next;
}s[maxn];
int head;int n,k;
vector<node>v;
int main(){
    int add;
    cin>>head>>n>>k;
    for(int i=0;i<n;i++){
        cin>>add;
         cin>>s[add].data>>s[add].next;
         s[add].address=add;
    }
    for(int i=head;i!=-1;i=s[i].next) v.push_back(s[i]);
    for(int i=1;i<v.size();i++){
        int left=(i-1)*k,right=left+k;
        if(right>v.size()) break;
        reverse(v.begin()+left,v.begin()+right);
    }
    for(int i=0;i<v.size();i++){
        printf("%05d %d ",v[i].address,v[i].data);
//        cout<<t[i].address<<" "<<t[i].data<<" ";
        if(i!=v.size()-1) printf("%05d\n",v[i+1].address);
        else cout<<"-1"<<endl;
    }
}