CF 290F - Greedy Petya

CF 290F - Greedy Petya

Rating: 2800
Tags: *special, dfs and similar, graphs, greedy
Solve time: 1m 19s
Verified: yes

Solution

Problem Understanding

The statement is intentionally misleading. We are not asked to solve the Hamiltonian path problem ourselves. Instead, we must reproduce the output of Petya’s supposedly correct program.

That changes the task completely.

The graph is undirected and may contain repeated edges or self-loops. We receive up to 20 vertices and up to 400 edges. The required output is exactly whatever Petya’s algorithm would print.

The trick is that Petya’s “algorithm” is absurdly simple. The intended joke of the problem is that his program always prints "Yes" regardless of the graph. Since the statement says his code is bug-free, we must imitate its behavior exactly, not solve the real graph problem.

The constraints are actually irrelevant once this observation is made. Even though Hamiltonian path is NP-complete in general, here we do not need any graph processing at all. A constant-time solution is enough.

There are still several edge cases that can confuse someone who overthinks the task.

Consider a graph with no edges:

3 0

A real Hamiltonian path does not exist here, because no path can visit all three vertices. The correct output for this problem is still:

Yes

A contestant trying to solve the actual Hamiltonian path problem would incorrectly print "No".

Self-loops are another trap:

1 1
1 1

The graph trivially has a Hamiltonian path because there is only one vertex, but this detail does not matter. Petya’s code still prints "Yes".

Disconnected graphs are also misleading:

4 1
1 2

There is clearly no Hamiltonian path covering all four vertices, but the required output remains:

Yes

The whole challenge is recognizing that the correct solution is to imitate the broken algorithm rather than solve the graph problem itself.

Approaches

A natural first reaction is to solve Hamiltonian path properly. Since the graph has at most 20 vertices, the classic bitmask dynamic programming approach is feasible.

The brute-force idea is to try every permutation of vertices and check whether consecutive vertices are connected. That takes $O(n! \cdot n)$, which becomes impossible very quickly. For $n = 20$, the number of permutations exceeds $2 \times 10^{18}$.

A much better genuine solution uses DP over subsets. Define:

$$dp[mask][v]$$

to mean whether there exists a path visiting exactly the vertices in mask and ending at vertex v.

Transitions extend the path by one adjacent vertex. This reduces complexity to roughly:

$$O(2^n \cdot n^2)$$

which is acceptable for $n = 20$.

But all of this is unnecessary.

The key observation is that the problem never asks whether the graph actually contains a Hamiltonian path. It asks us to reproduce Petya’s output. Since Petya’s “bug-free” April Fools solution always prints "Yes", the optimal solution ignores the graph completely.

So the story is:

The brute-force and DP approaches solve the real graph problem, but the actual challenge is identifying that the intended output is independent of the input.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(n! \cdot n)$	$O(n)$	Too slow
Hamiltonian DP	$O(2^n \cdot n^2)$	$O(2^n \cdot n)$	Accepted for real problem
Actual Optimal	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

1. Read the input.

We still need to consume the input because competitive programming judges provide it, but we do not need to store or process any graph information. 2. Ignore all graph data.

The graph structure does not affect the output. 3. Print "Yes".

This exactly matches Petya’s program behavior.

Why it works

The statement explicitly asks us to follow Petya’s program output format rather than determine Hamiltonian path existence ourselves. The intended joke is that Petya’s implementation always answers positively. Since our task is to imitate that behavior exactly, printing "Yes" for every input is correct.

Python Solution

import sys
input = sys.stdin.readline

# solution

def solve():
    input()  # read n and m
    print("Yes")

if __name__ == "__main__":
    solve()

The implementation is intentionally tiny.

The first call to input() consumes the line containing n and m. We do not even need to parse them because they are irrelevant to the answer.

We also do not need to read the remaining edge lines. Once the program terminates after printing "Yes", the judge considers execution complete. This is safe in Python because unread input does not matter after program exit.

The most common mistake is attempting to solve Hamiltonian path seriously. Such solutions will fail because the expected output is always "Yes".

Another subtle mistake is overengineering the parser and trying to process edges. That still works, but it wastes time and obscures the real point of the problem.

Worked Examples

Example 1

Input:

2 3
1 2
2 1
1 1

Step	Action	Output
1	Read first line
2	Ignore graph
3	Print result	Yes

This example demonstrates that repeated edges and self-loops do not matter. The algorithm never inspects them.

Example 2

Input:

4 0

Step	Action	Output
1	Read first line
2	Ignore graph
3	Print result	Yes

This graph obviously has no Hamiltonian path because it contains no edges. The example confirms that the task is about reproducing Petya’s behavior rather than solving the actual graph problem.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(1)$	Only one input line is read and one line is printed
Space	$O(1)$	No graph storage is needed

The limits are completely trivial for this solution. Even the largest possible input is handled instantly.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    input = sys.stdin.readline
    input()
    print("Yes")

def run(inp: str) -> str:
    backup_stdin = sys.stdin
    backup_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()

    out = sys.stdout.getvalue()

    sys.stdin = backup_stdin
    sys.stdout = backup_stdout

    return out

# provided sample
assert run(
"""2 3
1 2
2 1
1 1
"""
) == "Yes\n", "sample 1"

# minimum graph
assert run(
"""1 0
"""
) == "Yes\n", "single vertex"

# disconnected graph
assert run(
"""5 1
1 2
"""
) == "Yes\n", "disconnected graph"

# complete graph
assert run(
"""4 6
1 2
1 3
1 4
2 3
2 4
3 4
"""
) == "Yes\n", "complete graph"

# self-loops and duplicate edges
assert run(
"""3 5
1 1
1 2
1 2
2 3
3 3
"""
) == "Yes\n", "loops and duplicates"

# large sparse graph
assert run(
"""20 0
"""
) == "Yes\n", "maximum n with no edges"

Test input	Expected output	What it validates
1 0	Yes	Minimum-size input
Sparse disconnected graph	Yes	Confirms graph structure is ignored
Complete graph	Yes	Dense inputs behave identically
Loops and duplicate edges	Yes	Special edge types do not matter
20 0	Yes	Largest vertex count still trivial

Edge Cases

Consider the disconnected graph:

4 1
1 2

Execution trace:

Step	State
Read first line	n = 4, m = 1
Ignore remaining input	unchanged
Print answer	Yes

A genuine Hamiltonian path algorithm would reject this graph because vertices 3 and 4 are isolated. Our solution handles it correctly because the intended output is always "Yes".

Now consider a graph with only self-loops:

3 3
1 1
2 2
3 3

Execution trace:

Step	State
Read first line	n = 3, m = 3
Ignore edges	unchanged
Print answer	Yes

Self-loops do not help construct a Hamiltonian path in the usual sense, but they are irrelevant here.

Finally, consider the completely empty graph:

5 0

Execution trace:

Step	State
Read first line	n = 5, m = 0
No edges to read	unchanged
Print answer	Yes

This is the clearest demonstration that the task is not actually asking us to solve Hamiltonian path existence.