CF 2055A - Two Frogs

Rating: 800
Tags: constructive algorithms, games, greedy, math
Solve time: 1m 26s
Verified: yes

Solution

Problem Understanding

We are asked to analyze a two-player game played on a row of lilypads numbered from 1 to $n$. Alice starts on lilypad $a$ and Bob on lilypad $b$, and they take turns jumping either one step left or right. A frog loses if it cannot jump to a valid, unoccupied lilypad. Alice moves first, and the question is whether she can guarantee a win if both play optimally.

The input consists of multiple test cases. Each test case gives $n$, the number of lilypads, and the starting positions $a$ and $b$. The output is "YES" if Alice can always win, "NO" otherwise.

Because $n$ is at most 100, any algorithm that performs a small number of operations per position is feasible. The low bound suggests that even thinking about positions sequentially is acceptable. Edge cases arise when a frog starts near the edge (positions 1 or $n$) or when the frogs are adjacent. For instance, if Alice is at 1 and Bob is at 2, Alice cannot move left, and moving right lands next to Bob, potentially creating a losing position. If Alice cannot jump anywhere, she immediately loses.

A naive approach might try simulating all possible sequences of jumps recursively. This could work for small $n$, but it is unnecessary because the game is symmetric and the optimal moves can be derived from the distance to the edges.

Approaches

The brute-force approach is to simulate the game recursively: for each frog's turn, try every valid move, then recurse to the next turn. For each game state defined by Alice and Bob's positions, we check if the current player has a move that leads to a win. This is correct because it explores every legal sequence, but the number of states is $O(n^2)$ (every pair of positions) and each recursion branches into up to 2 moves. This could result in hundreds of recursive calls per test case and would be overkill for 500 test cases.

The key insight is that the game boils down to whether Alice can push Bob to the edge first. If the number of lilypads to the nearest edge from Alice is smaller than that from Bob, she can reach the edge faster and force Bob into a position with no moves. Formally, Alice can win if the minimum distance from her position to the nearest end of the row is strictly less than Bob's minimum distance to the nearest end. If both distances are equal, Bob can mirror Alice's moves and win instead.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^(n^2))	O(n^2)	Too slow
Optimal	O(1) per test case	O(1)	Accepted

Algorithm Walkthrough

Compute Alice's distance to the nearest edge: $\text{dA} = \min(a-1, n-a)$. This represents how quickly Alice can reach a lilypad where Bob cannot block her.
Compute Bob's distance to the nearest edge: $\text{dB} = \min(b-1, n-b)$. This is Bob's analogous distance.
Compare the two distances. If $\text{dA} < \text{dB}$, Alice can reach the edge first and guarantee a win. Print "YES".
Otherwise, Alice cannot guarantee a win because Bob can either reach an edge first or mirror her moves to block her. Print "NO".

The invariant is that the player closer to an edge has control of the game. By always moving toward that edge, Alice can reduce Bob’s options until he is trapped. No other sequence of moves improves her position if she is not closer to an edge than Bob.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, a, b = map(int, input().split())
    # compute distance to nearest edge
    dA = min(a - 1, n - a)
    dB = min(b - 1, n - b)
    if dA < dB:
        print("YES")
    else:
        print("NO")

We first read the number of test cases. For each test case, we read the lilypad count and frog positions. The distances to edges are computed directly and compared. Boundary conditions are correctly handled because min(a-1, n-a) always yields a non-negative integer representing steps to the closest end. Since $a \neq b$ and both are within bounds, no additional checks are necessary.

Worked Examples

Sample Input 1:

n	a	b	dA	dB	Alice Wins?
2	1	2	0	1	NO
3	3	1	0	0	YES
4	2	3	1	1	NO
5	2	4	1	1	YES
7	6	2	1	1	YES

In the first row, Alice is at the edge (distance 0), Bob has distance 1. Alice cannot move to a safe spot, so she loses. The table demonstrates how the distance comparison captures the winning condition correctly.

Sample Input 2:

n	a	b	dA	dB	Alice Wins?
5	1	5	0	0	NO
6	2	3	1	2	YES
10	4	7	3	3	NO

This shows that even if Alice is closer to one edge, if Bob is equally close to his edge, Alice cannot force a win.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case requires only a constant number of arithmetic operations and a comparison.
Space	O(1)	We only store a few integers per test case; no extra memory is required.

Given $t \le 500$ and $n \le 100$, the solution executes in milliseconds and uses negligible memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    t = int(input())
    for _ in range(t):
        n, a, b = map(int, input().split())
        dA = min(a - 1, n - a)
        dB = min(b - 1, n - b)
        print("YES" if dA < dB else "NO")
    return out.getvalue().strip()

# Provided samples
assert run("5\n2 1 2\n3 3 1\n4 2 3\n5 2 4\n7 6 2\n") == "NO\nYES\nNO\nYES\nYES", "Sample 1"

# Custom test cases
assert run("3\n5 1 5\n6 2 3\n10 4 7\n") == "NO\nYES\nNO", "Edge distance cases"
assert run("2\n2 1 2\n2 2 1\n") == "NO\nNO", "Minimum n, adjacent positions"
assert run("1\n100 50 51\n") == "YES", "Middle positions, Alice closer to left edge"
assert run("1\n100 51 50\n") == "NO", "Middle positions, Bob closer to left edge"

Test input	Expected output	What it validates
`5 1 5\n6 2 3\n10 4 7`	NO, YES, NO	Edge distances and mid positions
`2 1 2\n2 2 1`	NO, NO	Minimum n and adjacent positions
`100 50 51`	YES	Alice closer to nearest edge in middle of row
`100 51 50`	NO	Bob closer to nearest edge in middle of row

Edge Cases

If Alice starts at position 1 and Bob at 2 in a 2-lilypad game, Alice cannot jump left and moving right lands adjacent to Bob. The algorithm computes dA = min(1-1, 2-1) = 0 and dB = min(2-1, 2-2) = 0. Since dA < dB is false, the output is NO, which correctly identifies that Alice loses immediately. Similarly, if Alice and Bob start in the middle with equal distance to their respective edges, the algorithm outputs NO, capturing the symmetric mirroring scenario. The distance comparison fully handles edge adjacency, middle positions, and minimal board sizes without additional branching logic.