The 2020 US Open Platinum contest was a challenging event for young programmers. Problem 2, "Exercise", presented a particularly intriguing puzzle. In this article, we'll dive into the problem statement, analyze its solution, and provide insights into the key concepts involved.
Understanding the Problem
Problem 2: Exercise asks you to determine the minimum number of exercises a cow can complete in a given timeframe, given certain constraints. The problem revolves around a group of cows each having a unique exercise duration and a starting time.
Here's the breakdown:
- Input: You are given a list of
Ncows, where each cow has astarttime and adurationfor their exercise. - Constraints: The cows can only start their exercises at the given
starttimes, and they cannot switch exercises. The goal is to maximize the number of cows that can complete their exercises within a given time frameT.
Solution Approach
The key to solving this problem lies in understanding that the most efficient strategy is to prioritize cows with shorter exercise durations. This approach maximizes the likelihood of completing more exercises within the time limit.
Here's how we can implement this:
- Sorting: First, sort the cows in ascending order of their
duration. This ensures that cows with shorter exercises are considered first. - Iteration: Iterate through the sorted list of cows. For each cow, check if its
starttime plus itsdurationis less than or equal to the time limitT. If so, increment the count of completed exercises and move on to the next cow. Otherwise, skip the cow as it cannot be completed within the time limit.
Code Implementation
#include
#include
#include
using namespace std;
struct Cow {
int start;
int duration;
};
bool compareCows(const Cow& a, const Cow& b) {
return a.duration < b.duration;
}
int main() {
int N, T;
cin >> N >> T;
vector cows(N);
for (int i = 0; i < N; i++) {
cin >> cows[i].start >> cows[i].duration;
}
sort(cows.begin(), cows.end(), compareCows);
int completedExercises = 0;
for (int i = 0; i < N; i++) {
if (cows[i].start + cows[i].duration <= T) {
completedExercises++;
} else {
break; // No more cows can complete within the time limit
}
}
cout << completedExercises << endl;
return 0;
}
Example
Let's consider a scenario with:
- Time Limit (T): 10
- Cows:
- Cow 1:
start = 2,duration = 5 - Cow 2:
start = 1,duration = 3 - Cow 3:
start = 4,duration = 6 - Cow 4:
start = 0,duration = 2
- Cow 1:
After sorting by duration, the order becomes:
- Cow 4:
start = 0,duration = 2 - Cow 2:
start = 1,duration = 3 - Cow 1:
start = 2,duration = 5 - Cow 3:
start = 4,duration = 6
Following the algorithm:
- Cow 4:
start + duration = 2(Completes within the time limit) - Cow 2:
start + duration = 4(Completes within the time limit) - Cow 1:
start + duration = 7(Completes within the time limit) - Cow 3:
start + duration = 10(Completes within the time limit)
Therefore, the maximum number of exercises that can be completed within the time limit is 4.
Conclusion
The "Exercise" problem from the 2020 US Open Platinum contest provides a classic example of how a greedy algorithm can be effectively applied to find the optimal solution. By prioritizing cows with shorter exercise durations, we ensure the maximum number of exercises are completed within the time limit. This approach highlights the importance of understanding problem constraints and applying efficient strategies to solve them.
