U. Toy Train Tracks

46th Annual hosted by ICPC World Championship AASTMT Problem U Toy Train Trac...

Updated May 21, 2026

Track ICPC

Year 2022

Statement Text + PDF

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Problem statement

Problem Statement

Formatted from the contest statement text, with sample tests broken out into copyable blocks.

Statement text Statement PDF

   46th Annual                                                                                          hosted by
  ICPC World
Championship                                                                                            AASTMT

Problem U Toy Train Tracks Time limit: 1 second Every little child, and quite a number of adults, are fascinated by toy trains. From a toddler’s choo-choo train to a hobbyist’s elabo- rate model railroad filling an entire basement, they are a profitable business. The Toy Train Tracks Construction Company (TTTCC) manufactures train tracks for all ages and skill levels. To keep their existing customers busy and maybe attract some new ones, the TTTCC has recently started publishing maps for how to con- nect their train tracks into elaborate layouts. Usually, this starts with a designer coming up with an interesting track layout, and then publishing both the layout and the required number of dif- Model railroad by Marshman via Wikimedia Commons, cc by-sa ferent track segments (say, curves and straight parts) needed to construct it. But the TTTCC has recently learned that many customers are looking for the reverse: they already have train track segments lying around (maybe found in grandma’s attic), and would like to use them to create a large train course. How difficult might that be? To study the feasibility of automating the layout-creation process, TTTCC is interested in constructing train courses using two different shapes: straight line segments, and 90-degree turns (see Figure U.1).

Figure U.1: A straight track segment and a curved track segment.

Valid layouts are created by placing these shapes on a square grid, with each track piece taking up exactly one grid cell. Both types of pieces can be rotated in 90-degree increments. A “proper” train track needs to be connected, and should form a single closed loop. Given a set of straight and curved track segments, what is the longest closed loop that one can construct?

Figure U.2: A sample track using four straight track segments and twelve curves. This
corresponds to Sample Output 1.

                                     World Finals | ICPC 2023 Luxor
  46th Annual                                                                               hosted by
  ICPC World
Championship                                                                                AASTMT

Input

The input consists of a single line containing two integers s and c, the number of straight and curved track segments available, respectively (0 ≤ s ≤ 105 , 4 ≤ c ≤ 105 ).

Output

Output a train loop using at most s straight segments and c curved segments, that has the longest length (in number of track segments used) under this restriction. The loop must be closed and cannot intersect itself. If there are multiple loops of maximal length, any one of them will be accepted. If the loop is of length n, then print a single string of length n, where the characters represent the loop’s segments as encountered in a single traversal. The character ‘S‘ stands for a straight-line segment, ‘L‘ for a curved segment that is a left turn, and ‘R’ for a curved segment that is a right turn.

Sample Tests

Sample 1

Sample Input

 4 12

Sample Output

LSRLLRLSLSRLLSRL

Sample 2

Sample Input

1 5

Sample Output

LLLL

Editorial

The solution write-up is rendered from the LaTeX source, with equations kept live through MathJax.

Raw TeX

Key Observations

In every closed orthogonal loop, the number of straight pieces is even and the number of curved pieces is even. So we may discard one unused piece of either type if its count is odd.
If no straight pieces are used, the best possible lengths are:
- $4$, using LLLL;
- or any multiple of $4$ at least $12$, using a four-sided staircase loop.
- There is no all-curves loop of length $8$.
- If at least two straight pieces are available, then every even number of curves at least $4$ is achievable together with every even number of straight pieces.
- The implementation constructs the loop as a sequence of absolute movement directions on the grid and converts that cyclic direction sequence into the required alphabet S/L/R by comparing consecutive directions.
Algorithm
1. Replace $(s,c)$ by the largest even counts not exceeding them.
2. If fewer than two straight pieces remain:
  - use $4$ curves if fewer than $12$ curves remain;
  - otherwise use the largest multiple of $4$ not exceeding $c$.
  - If at least two straight pieces remain and $c \equiv 0 \pmod 4$, build the direction sequence \[ N^a\ W\ (SW)^q\ S^a\ E\ (NE)^q, \] where $a=s/2+1$ and $q=(c-4)/4$.
  - If at least two straight pieces remain and $c \equiv 2 \pmod 4$, build \[ N^2\ (WN)^p\ W^b\ (SE)^{p+1}\ S\ E^{b-1}, \] where $b=s/2+1$ and $p=(c-6)/4$.
  - Convert the cyclic direction sequence into S/L/R and output it. enumerate
    
    Correctness Proof
    
    We prove that the algorithm returns the correct answer.
    
    Lemma 1.
    
    Any closed loop uses an even number of straight pieces and an even number of curved pieces.
    
    Proof.
    
    The total horizontal displacement and the total vertical displacement of a closed loop are both zero, so the number of steps in each direction must balance. This forces the number of straight pieces to be even. Also, the total turning angle of a simple closed loop is $\pm 360^\circ$, so the difference between the numbers of left and right turns is $\pm 4$. Therefore their sum, which is the total number of curved pieces, is even. □
    
    Lemma 2.
    
    The direction families used by the algorithm are simple closed loops and use exactly the claimed numbers of straight and curved pieces.
    
    Proof.
    
    Each family is a concatenation of monotone staircase sides. Their horizontal and vertical displacements cancel by construction, so the walk closes. Because the sides are monotone and nested outward, the walk never revisits a cell before the end, so it is simple.
    
    Counting consecutive equal directions gives the number of straight pieces, and counting direction changes gives the number of curved pieces. For the two families these counts are respectively \[ 2a-2,\ 4+4q \quad\text{and}\quad 2b-2,\ 6+4p, \] which are exactly the requested values after substituting $a=s/2+1$, $b=s/2+1$, $q=(c-4)/4$, and $p=(c-6)/4$. □
    
    Lemma 3.
    
    For the all-curves case, the algorithm uses the largest feasible number of curves.
    
    Proof.
    
    By Lemma 1, only even numbers of curves are possible. The staircase all-curves family realizes every multiple of $4$ at least $12$, and LLLL realizes $4$. The only even value below $12$ not covered by these constructions is $8$, which is impossible. Hence the algorithm picks the largest feasible number of curves when no straight pieces can be used. □
    
    Theorem.
    
    The algorithm outputs a loop of maximum possible length.
    
    Proof.
    
    By Lemma 1, any optimal solution can use only even numbers of straight and curved pieces, so rounding the available counts down to even values loses nothing.
    
    If fewer than two straight pieces remain, Lemma 3 shows that the algorithm uses the maximum possible number of curves. Otherwise, Lemma 2 shows that the algorithm can realize every even curve count at least $4$ together with every even straight count, so it uses all available pieces after the parity adjustment. No longer loop can exist. □
    
    Complexity Analysis
    
    The algorithm constructs the answer string directly. The running time is linear in the output length, and the memory usage is also linear in the output length.
    
    Implementation Notes
    - The output does not need to match the sample exactly; any maximum-length simple loop is accepted.

C++

Clean code view with a raw-file link when you want the original source.

Raw file

#include <bits/stdc++.h>
using namespace std;

namespace {

vector<int> build_all_curves(int curves) {
    if (curves == 4) {
        return {1, 2, 3, 0};  // LLLL
    }

    int a = 2;
    int b = curves / 4 - 2;
    vector<int> dirs;
    dirs.reserve(curves);
    for (int i = 0; i < a; ++i) {
        dirs.push_back(1);
        dirs.push_back(0);
    }
    for (int i = 0; i < b; ++i) {
        dirs.push_back(3);
        dirs.push_back(0);
    }
    for (int i = 0; i < a; ++i) {
        dirs.push_back(3);
        dirs.push_back(2);
    }
    for (int i = 0; i < b; ++i) {
        dirs.push_back(1);
        dirs.push_back(2);
    }
    return dirs;
}

vector<int> build_mod0(int straights, int curves) {
    int a = straights / 2 + 1;
    int q = (curves - 4) / 4;

    vector<int> dirs;
    dirs.reserve(straights + curves);
    dirs.insert(dirs.end(), a, 1);  // N^a
    dirs.push_back(2);              // W
    for (int i = 0; i < q; ++i) {
        dirs.push_back(3);          // S
        dirs.push_back(2);          // W
    }
    dirs.insert(dirs.end(), a, 3);  // S^a
    dirs.push_back(0);              // E
    for (int i = 0; i < q; ++i) {
        dirs.push_back(1);          // N
        dirs.push_back(0);          // E
    }
    return dirs;
}

vector<int> build_mod2(int straights, int curves) {
    int a = 2;
    int b = straights / 2 + 1;
    int p = (curves - 6) / 4;

    vector<int> dirs;
    dirs.reserve(straights + curves);
    dirs.insert(dirs.end(), a, 1);  // N^2
    for (int i = 0; i < p; ++i) {
        dirs.push_back(2);          // W
        dirs.push_back(1);          // N
    }
    dirs.insert(dirs.end(), b, 2);  // W^b
    for (int i = 0; i < p + 1; ++i) {
        dirs.push_back(3);          // S
        dirs.push_back(0);          // E
    }
    dirs.push_back(3);              // S
    dirs.insert(dirs.end(), b - 1, 0);  // E^(b-1)
    return dirs;
}

string directions_to_tracks(const vector<int>& dirs) {
    string answer;
    answer.reserve(dirs.size());

    int prev = dirs.back();
    for (int dir : dirs) {
        if (dir == prev) {
            answer.push_back('S');
        } else if (dir == (prev + 1) % 4) {
            answer.push_back('L');
        } else {
            answer.push_back('R');
        }
        prev = dir;
    }
    return answer;
}

void solve() {
    int s, c;
    cin >> s >> c;

    int use_s = s - (s & 1);
    int use_c = c - (c & 1);

    vector<int> dirs;
    if (use_s < 2) {
        int use_curves;
        if (use_c < 12) {
            use_curves = 4;
        } else {
            use_curves = use_c - (use_c % 4);
        }
        dirs = build_all_curves(use_curves);
    } else if (use_c % 4 == 0) {
        dirs = build_mod0(use_s, use_c);
    } else {
        dirs = build_mod2(use_s, use_c);
    }

    cout << directions_to_tracks(dirs) << '\n';
}

}  // namespace

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();
    return 0;
}

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TeX write-upcompetitive_programming/icpc/2022/U-toy-train-tracks/solution.texC++ implementationcompetitive_programming/icpc/2022/U-toy-train-tracks/solution.cppStatement textcompetitive_programming/icpc/2022/U-toy-train-tracks/statement.txtStatement PDFcompetitive_programming/icpc/2022/U-toy-train-tracks/statement.pdfMetadatacompetitive_programming/icpc/2022/U-toy-train-tracks/meta.jsonYear packetcompetitive_programming/icpc/2022/contest_problems.pdf

U. Toy Train Tracks

Problem Statement

Input

Output

Sample Tests

Editorial

Key Observations

Algorithm

Correctness Proof

Complexity Analysis

Implementation Notes

Code

Source Files and Assets