摘要
Title: 4273. 链表合并
Tag: 链表
Memory Limit: 64 MB
Time Limit: 1000 ms

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• 题意
• 思路
• 代码

4273. 链表合并

• 题意

  见原题

• 思路

  翻转较短的链表后，挨个插入即可

• 代码

  /*
  * @Author: NEFU AB-IN
  * @Date: 2022-06-20 21:45:36
  * @FilePath: \ACM\Acwing\4273.cpp
  * @LastEditTime: 2022-06-20 22:12:55
  */
  #include <bits/stdc++.h>
  using namespace std;
  #define SZ(X) ((int)(X).size())
  #define IOS                                                                                                            \
      ios::sync_with_stdio(false);                                                                                       \
      cin.tie(0);                                                                                                        \
      cout.tie(0);
  #define DEBUG(X) cout << #X << ": " << X << endl;
  typedef pair<int, int> PII;
  
  const int N = 1e6 + 10;
  
  int e[N], ne[N];
  
  signed main()
  {
      int h1, h2, n;
      cin >> h1 >> h2 >> n;
  
      for (int i = 1; i <= n; ++i)
      {
          int addr, d, nxt;
          cin >> addr >> d >> nxt;
          e[addr] = d;
          ne[addr] = nxt;
      }
  
      // 统计结点
      vector<int> v1, v2;
      for (int i = h1; ~i; i = ne[i])
      {
          v1.push_back(i);
      }
      for (int i = h2; ~i; i = ne[i])
      {
          v2.push_back(i);
      }
  
      if (SZ(v1) > SZ(v2))
          swap(v1, v2);
  
      vector<int> ans;
      reverse(v1.begin(), v1.end());
  
      for (int i = 0, j = 0; i < SZ(v2); ++i)
      {
          ans.push_back(v2[i]);
          if (i & 1 && j < SZ(v1))
              ans.push_back(v1[j++]);
      }
  
      for (int i = 0; i < SZ(ans); ++i)
      {
          if (i < SZ(ans) - 1)
              printf("%05d %d %05d\n", ans[i], e[ans[i]], ans[i + 1]);
          else
              printf("%05d %d -1", ans[i], e[ans[i]]);
      }
      return 0;
  }