G. Pipe Stream
Your hometown has hired some contractors – including you! – to man- age its municipal pipe network. They built the network, at great ex- pense, to supply Flubber to every home in town. Unfortunately, no- body has found a use for Flubber yet, but never mind....
Problem Statement
Formatted from the contest statement text, with sample tests broken out into copyable blocks.
Your hometown has hired some contractors – including you! – to man- age its municipal pipe network. They built the network, at great ex- pense, to supply Flubber to every home in town. Unfortunately, no- body has found a use for Flubber yet, but never mind. It was a Flubber network or a fire department, and honestly, houses burn down so rarely, a fire department hardly seems necessary. In the possible event that somebody somewhere decides they want some Flubber, they would like to know how quickly it will flow through the pipes. Measuring its rate of flow is your job. You have access to one of the pipes connected to the network. The pipe is l meters long, and you can start the flow of Flubber through this pipe at a time of your choosing. You know that it flows with a constant real-valued speed, which is at least v1 meters/second and at most v2 Picture by Nevit via Wikimedia Commons meters/second. You want to estimate this speed with an absolute error of at most 2t meters/second. Unfortunately, the pipe is opaque, so the only thing you can do is to knock on the pipe at any point along its length, that is, in the closed real-valued range [0, l]. Listening to the sound of the knock will tell you whether or not the Flubber has reached that point. You are not infinitely fast. Your first knock must be at least s seconds after starting the flow, and there must be at least s seconds between knocks. Determine a strategy that will require the fewest knocks, in the worst case, to estimate how fast the Flubber is flowing. Note that in some cases the desired estimation might be impossible (for example, if the Flubber reaches the end of the pipe too quickly).
Input
The input consists of multiple test cases. The first line of input contains an integer c (1 ≤ c ≤ 100), the number of test cases. Each of the next c lines describes one test case. Each test case contains the five integers l, v1 , v2 , t and s (1 ≤ l, v1 , v2 , t, s ≤ 109 and v1 < v2 ), which are described above.
Output
For each test case, display the minimal number of knocks required to estimate the flow speed in the worst case. If it might be impossible to measure the flow speed accurately enough, display impossible instead.
Sample Tests
3
1000 1 30 1 1
60 2 10 2 5
59 2 10 2 5 5
3
impossible Editorial
The solution write-up is rendered from the LaTeX source, with equations kept live through MathJax.
Key Observations
Write the structural observations that make the problem tractable.
State any useful invariant, monotonicity property, graph interpretation, or combinatorial reformulation.
If the constraints matter, explain exactly which part of the solution they enable.
Algorithm
Describe the data structures and the state maintained by the algorithm.
Explain the processing order and why it is sufficient.
Mention corner cases explicitly if they affect the implementation.
Correctness Proof
We prove that the algorithm returns the correct answer.
Lemma 1.
State the first key claim.
Proof.
Provide a concise proof.
Lemma 2.
State the next claim if needed.
Proof.
Provide a concise proof.
Theorem.
The algorithm outputs the correct answer for every valid input.
Proof.
Combine the lemmas and finish the argument.
Complexity Analysis
State the running time and memory usage in terms of the input size.
Implementation Notes
Mention any non-obvious implementation detail that is easy to get wrong.
Mention numeric limits, indexing conventions, or tie-breaking rules if relevant.
Code
C++ solution used for this page.
#include <bits/stdc++.h>
using namespace std;
namespace {
void solve() {
// Fill in the full solution logic for the problem here.
}
} // namespace
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
solve();
return 0;
}
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