1074 Reversing Linked List (25 分)

反转子链表
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.
每隔固定长度对链表进行一次翻转,最后按题目要求输出

  • 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 
#include 
#include 
#include 
#include 
using namespace std;
struct data{
    int val;
    int next;
};
vector v(100000);
vector v_n2a;
int main(){
    int n,m,start;
    scanf("%d %d %d", &start,&n, &m);
    int address, val, next;
    int i, j, k;
    for(i=0;i

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