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 <iostream>
#include <vector>
#include <map>
#include <string>
#include <algorithm>
using namespace std;
struct data{
    int val;
    int next;
};
vector<data> v(100000);
vector<data> 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<n;i++){
        scanf("%d %d %d",&address, &val, &next);
        v[address].val = val;
        v[address].next = next;
    }
    for(i=start;i!=-1;i=v[i].next){
        //printf("%05d %d %05d\n",i,v[i].val,v[i].next);
        v_n2a.push_back({v[i].val, i});
    }
    for(i=0;i+m-1<v_n2a.size();i+=m){
        j = i+m-1;
        k = i;
        while(k<j){
            //cout << i << " " <<v_n2a[i].val << endl;
            swap(v_n2a[k],v_n2a[j]);
            //cout << i << " " << v_n2a[i].val << endl;
            k++,j--;
        }
    }
    for(i=0;i<v_n2a.size();i++){
        if(i!=v_n2a.size()-1){
            printf("%05d %d %05d\n",v_n2a[i].next,v_n2a[i].val,v_n2a[i+1].next);
        }else{
            printf("%05d %d -1\n",v_n2a[i].next,v_n2a[i].val);
        }
    }
    return 0;
}