Solving AT_abc429_d: A Detailed Explanation

This article provides an in-depth explanation of the solution to the AT_abc429_d problem. We will break down the problem, discuss the key concepts, and walk through the code implementation. Whether you're a beginner or an experienced programmer, this guide will help you understand the solution and improve your problem-solving skills.

Understanding the Problem

Before diving into the solution, let's first understand the problem statement. We are given a large pond's circumference and a limited number of people standing at different positions around the pond. The goal is to calculate the sum of the number of people within a certain range for each person.

The problem's constraints make a naive $\mathcal{O}(n^2)$ solution infeasible. Therefore, we need to find a more efficient approach. The key observation is that the circumference is large, but the number of people is limited. This suggests that we can discretize the positions and count the people at each unique location.

Key Concepts and Algorithm

The core idea behind the solution is to use a two-pointer technique to efficiently calculate the number of people within the required range for each person. Here's a breakdown of the steps involved:

1. Discretization and Counting:

   • First, we discretize the positions of the people. This means we identify the unique positions around the pond and store them.
   • We use a map (or a similar data structure) to count the number of people at each unique position. This helps us avoid redundant calculations later on. The discretization process significantly reduces the problem's complexity by focusing on unique locations rather than individual positions.

2. Sorting:

   • Next, we sort the unique positions in ascending order. This step is crucial for the two-pointer technique to work efficiently. Sorting enables us to iterate through the positions in a structured manner, making it easier to track the range of people we need to consider. Sorting is a fundamental step in many algorithms, particularly when dealing with ranges and intervals.

3. Two-Pointer Technique:

   • The heart of the solution lies in the two-pointer technique. We use two pointers, l (left) and r (right), to define a range around each person's position.
   • For each person at position l, we expand the range by moving the right pointer r until the number of people within the range is greater than or equal to the given threshold c.
   • The critical observation here is that as we move from position i to i+1, the range of people included in the count will largely overlap. This is because the requirement to have more than c people means that the right endpoint of the range will likely shift only slightly. This overlapping nature is what makes the two-pointer approach so efficient.

4. Calculating the Answer:

   • As we move the pointers, we maintain a count of the number of people within the range. For each position l, we multiply the count by the distance covered by that position and add it to the total answer. The distance calculation considers the circular nature of the pond, ensuring that we correctly account for ranges that wrap around the circumference.

Code Walkthrough

Let's examine the C++ code provided in the original solution:

#include <bits/stdc++.h>
using namespace std;
using LL = long long;

struct P { LL id, num; } p[500005];
LL n, m, c, a[500005], cnt, r, now, ans;
map<LL, LL> pl;

bool cmp(P x, P y) { return x.id < y.id; }

LL dis(LL l) {
    if (l == cnt) return p[1].id + m - p[l].id;
    return p[l + 1].id - p[l].id;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m >> c;
    for (LL i = 1; i <= n; i++) {
        cin >> a[i];
        if (pl.count(a[i])) p[pl[a[i]]].num++;
        else {
            cnt++;
            p[cnt].id = a[i];
            p[cnt].num = 1;
            pl[a[i]] = cnt;
        }
    }
    sort(p + 1, p + cnt + 1, cmp);
    r = 1; now = p[1].num;
    for (LL l = 1; l <= cnt; l++) {
        now -= p[l].num;
        while (now < c) {
            r++;
            if (r > cnt) r -= cnt;
            now += p[r].num;
        }
        ans += now * dis(l);
    }
    cout << ans;
}

Let's break down the code section by section:

• Includes and Definitions:

  #include <bits/stdc++.h>
  using namespace std;
  using LL = long long;
  
  struct P { LL id, num; } p[500005];
  LL n, m, c, a[500005], cnt, r, now, ans;
  map<LL, LL> pl;
  

  • This section includes necessary headers, defines LL as a shorthand for long long, and declares the variables we'll be using. The P struct is used to store the ID (position) and the number of people at that position. The map pl is used for discretization.

• Comparison Function:

  bool cmp(P x, P y) { return x.id < y.id; }
  

  • This function is used for sorting the positions based on their IDs.

• Distance Function:

  LL dis(LL l) {
      if (l == cnt) return p[1].id + m - p[l].id;
      return p[l + 1].id - p[l].id;
  }
  

  • The dis function calculates the distance between two adjacent positions. It handles the circular nature of the pond by considering the wrap-around case (when l is the last position).

• Main Function:

  int main() {
      ios::sync_with_stdio(false);
      cin.tie(nullptr);
      cin >> n >> m >> c;
      for (LL i = 1; i <= n; i++) {
          cin >> a[i];
          if (pl.count(a[i])) p[pl[a[i]]].num++;
          else {
              cnt++;
              p[cnt].id = a[i];
              p[cnt].num = 1;
              pl[a[i]] = cnt;
          }
      }
      sort(p + 1, p + cnt + 1, cmp);
      r = 1; now = p[1].num;
      for (LL l = 1; l <= cnt; l++) {
          now -= p[l].num;
          while (now < c) {
              r++;
              if (r > cnt) r -= cnt;
              now += p[r].num;
          }
          ans += now * dis(l);
      }
      cout << ans;
  }
  

  • The main function is where the magic happens. Let's break it down further:
    • Input:

      cin >> n >> m >> c;
      for (LL i = 1; i <= n; i++) {
          cin >> a[i];
          if (pl.count(a[i])) p[pl[a[i]]].num++;
          else {
              cnt++;
              p[cnt].id = a[i];
              p[cnt].num = 1;
              pl[a[i]] = cnt;
          }
      }
      

      • This part reads the input values (n, m, c) and the positions of the people (a[i]). It then performs the discretization and counting using the map pl and the p array.
    • Sorting:

      sort(p + 1, p + cnt + 1, cmp);
      

      • This line sorts the unique positions using the cmp function.
    • Two-Pointer and Calculation:

      r = 1; now = p[1].num;
      for (LL l = 1; l <= cnt; l++) {
          now -= p[l].num;
          while (now < c) {
              r++;
              if (r > cnt) r -= cnt;
              now += p[r].num;
          }
          ans += now * dis(l);
      }
      

      • This is the core of the solution. It initializes the right pointer r and the current count now. The outer loop iterates through each position l. The inner loop moves the right pointer r until the count now is greater than or equal to c. The answer is then updated by multiplying the count by the distance covered by position l.
    • Output:

      cout << ans;
      

      • Finally, the calculated answer is printed.

Optimizations and Considerations

• Input/Output Optimization: The lines ios::sync_with_stdio(false); and cin.tie(nullptr); are used to optimize input/output operations in C++, which can be crucial for performance in competitive programming.
• Data Structures: The choice of map for discretization is efficient due to its logarithmic time complexity for insertion and lookup. However, other data structures like unordered_map could be considered for potentially better performance in some cases.
• Modularity: The code could be further improved by breaking it down into smaller, more modular functions. This would enhance readability and maintainability.

Conclusion

In this article, we have provided a detailed explanation of the solution to the AT_abc429_d problem. We discussed the key concepts, the algorithm, and the code implementation. The two-pointer technique, combined with discretization, is a powerful tool for solving problems involving ranges and intervals. By understanding the underlying principles and the code structure, you can apply these techniques to solve similar problems in the future.

For further learning on algorithmic problem-solving techniques, consider exploring resources like Competitive Programming Handbook. This comprehensive guide offers valuable insights and strategies for tackling various coding challenges.