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
  
 1 #include <stdio.h>
 2 #include <map>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <string>
 6 #include <math.h>
 7 #include <vector>
 8 using namespace std;
 9 const int maxn = 100010;
10 struct node{
11     int addr;
12     int data;
13     int next;
14 }nodes[maxn];
15 vector<node> v;
16 int main(){
17     int st, n, k, count = 0;
18     cin >> st >> n >> k;
19     for (int i = 0; i < n; i++){
20         int start, data, next;
21         cin >> start >> data >> next;
22         nodes[start].addr = start;
23         nodes[start].data = data;
24         nodes[start].next = next;
25     }
26     while (st != -1){
27         v.push_back(nodes[st]);
28         st = nodes[st].next;
29         count++;
30     }
31     for (int i = 0; i+k <= count; i = i + k){
32         reverse(v.begin() + i, v.begin() + i + k);
33     }
34     for (int i = 0; i < count-1; i++){
35         v[i].next = v[i + 1].addr;
36     }
37     v[count-1].next = -1;
38     for (int i = 0;i < count-1; i++){
39         printf("%05d %d %05d\n", v[i].addr, v[i].data, v[i].next);
40         
41     }
42     printf("%05d %d %d\n", v[count - 1].addr, v[count - 1].data, v[count - 1].next);
43     system("pause");
44 }
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注意点：同B1025，链表题，都会要你重新排序输出，注意可以用vector实现链表，不要用静态数组，保存节点时一定要保存自己的地址。反转可以直接使用algorithm里的reverse