Written in front

• Thinking analysis
  • Understanding the meaning
    • Given a list and K, after traversing the list, the nodes < 0 will be output first, then the nodes in the range of 0-k will be output, and finally the nodes in the range of > k will be output.
  • Specific realization:
    • 1. All nodes are stored in structure {id, data, next}.
    • 2. Traverse the list, find out the nodes in the list and put them in the container v.
    • 3. The nodes are divided into three categories (--infinite, 0), [0,k], (k, + -infinite)], which are put into the capacitive ans in sequence by segment and output at last.
• There is a certain degree of difficulty in the subject.
  • Understanding the meaning of a question takes time (core question)
  • Achieving Thought is Smart

test case

• input:
  00100 9 10
  23333 10 27777
  00000 0 99999
  00100 18 12309
  68237 -6 23333
  33218 -4 00000
  48652 -2 -1
  99999 5 68237
  27777 11 48652
  12309 7 33218
  output:
  33218 -4 68237
  68237 -6 48652
  48652 -2 12309
  12309 7 00000
  00000 0 99999
  99999 5 23333
  23333 10 00100
  00100 18 27777
  27777 11 -1
  

ac code

• #include <iostream>
  #include <vector>
  using namespace std;
  struct node
  {
      int id, data, next;
  };
  int main()
  {
      int bgin, n, k, s, d, e;
      scanf("%d%d%d", &bgin, &n, &k);
      node a[100010];
  
      vector<node> v, ans;
      for (int i = 0; i < n; i++)
      {
          scanf("%d%d%d", &s, &d, &e);
          a[s] = {s, d, e};
      }
      // Save the initial list sequence into an array
      for (; bgin != -1; bgin = a[bgin].next)
          v.push_back(a[bgin]);
  
      // Processing less than 0 elements
      for (int i = 0; i < v.size(); i++)
          if (v[i].data < 0) ans.push_back(v[i]);
      // Processing greater than or equal to 0, less than k elements
      for (int i = 0; i < v.size(); i++)
          if (v[i].data >= 0 && v[i].data <= k) ans.push_back(v[i]);
      // Processing elements larger than k
      for (int i = 0; i < v.size(); i++)
          if (v[i].data > k) ans.push_back(v[i]);
  
      for (int i = 0; i < ans.size() - 1; i++)
          printf("%05d %d %05d\n", ans[i].id, ans[i].data, ans[i + 1].id);
  
      printf("%05d %d -1\n", ans[ans.size() - 1].id, ans[ans.size() - 1].data);
      return 0;
  }
  

Summary of Knowledge Points

• Structural body assignment
  • a[s] = {s, d, e};