Optimizing Problem 702 C on Codeforces: A Detailed Guide

When tackling problem 702 C on Codeforces, the initial impression may lead you to believe that a straightforward O(nm) solution is sufficient. However, with careful consideration, we can achieve an efficient solution that leverages binary search and greedy methods. This guide will walk you through the problem-solving process and offer insights into optimizing the solution for better performance.

Understanding the Problem

Given m towers and n cities, the goal is to find the minimum range R in which a tower must be located to serve all cities. Each city can be served by one of the towers, and the distance can be at most R. The problem simplifies to determining the smallest R such that each city is within this range of a tower.

Naive Approach Analysis

The naive approach might involve considering each city and checking up to two towers (the previous and the next one) for every city. This results in a time complexity of O(nm). However, for large inputs, this can quickly become inefficient, especially with up to 1e9 potential towers.

Efficient Solution Strategy

To optimize the solution, we can follow these steps:

Check Feasibility: Determine if a given radius R can serve all cities. This can be done in O(n log m) time by using binary search over each city. Binary Search for Minimum Radius: Since the smallest possible radius is 0 and the largest is 2e9, we can use binary search to find the minimal R where it is possible to serve all cities. The search space is thus reduced to O(log 2e9), significantly improving the overall time complexity.

Detailed Steps

Step 1: Check Feasibility

For a given radius R, check if every city can be covered by at least one tower. This can be done by iterating over each city and using binary search to find the nearest towers within the range [x-R, x R] for each city, where x is the city's coordinate.

Step 2: Binary Search for Optimal Radius

Once we have a function that can determine if a given radius R is feasible, we can use binary search to find the smallest R. The binary search range is from 0 to 2e9, resulting in a complexity of O(n log m log 2e9).

Pseudo-Code

def check_feasibility(r, cities, towers):
    for city in cities:
        nearest_tower  None
        for tower in towers:
            if abs(tower - city)  r:
                nearest_tower  tower
                break
        if nearest_tower is None:
            return False
    return True
low  0
high  2e9
while low  high:
    mid  (low   high   1) // 2
    if check_feasibility(mid, cities, towers):
        high  mid
    else:
        low  mid   1
print(low)

In the provided pseudocode, the check_feasibility function uses binary search to find if there is a tower within the given radius for each city. The main function performs binary search to find the minimum radius.

Example Walkthrough

Considering the example provided in the initial problem statement, with cities at -5, -2, 2, 4 and towers at -3 and 0:

The first city at -5 can only be served by tower -3, so the initial minimum radius is 2. The second city at -2 is served by tower -3 (1 unit away), so the minimum radius remains 2. The third city at 2 is served by tower 0 (2 units away), and the fourth city at 4 is also served by tower 0 (4 units away).

In this case, the minimum radius R that satisfies the condition for all cities is 2.

Conclusion and Further Reading

By combining binary search with the greedy method, you can efficiently solve problem 702 C on Codeforces. The approach leverages the fact that once a feasible radius is found, any larger radius is also feasible. This reduces the overall time complexity significantly, making it well-suited for larger inputs. For further reading, review the Codeforces editorial or explore more problems involving binary search and greedy algorithms.