PAT 1074 Reversing Linked List (链表转置)

1074 Reversing Linked List (25分)

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

 链表转置，借用了vector和reverse函数

#include <bits/stdc++.h>
#define Max 100005 
using namespace std;
struct Node{
  int addr, data, nex;
}node[Max];
int first, n, k; 
int main(){
  int a, b, c, ind = 0;
  scanf("%d%d%d", &first, &n, &k);
  for(int i = 0; i < n; i++){
    scanf("%d%d%d", &a, &b, &c);
    node[a].addr = a;
    node[a].data = b;
    node[a].nex = c;
  }
  vector<Node> vec;
  while(first != -1) {
    vec.push_back(node[first]);
    first = node[first].nex;
  } 
  while(ind + k <= vec.size()){
    reverse(vec.begin() + ind, vec.begin() + ind + k);
    ind += k;
  }
  printf("%05d %d ",vec[0].addr, vec[0].data);
  for(int i = 1; i < vec.size(); i++){  //nex就是下一个地址 
    printf("%05d\n%05d %d ",vec[i].addr, vec[i].addr, vec[i].data);
  }
  printf("-1\n");
  return 0;
}