U USACO Prep

2024 February Silver · All three problems

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Statements paraphrased; canonical source is usaco.org. Each problem links to the official problem viewer.

Problem 1 · Target Practice II

Official problem (cpid=1398)

Statement (paraphrased)

4N cows stand on the y-axis. There are N axis-aligned rectangular targets to the right. Each cow fires along a line with a given slope toward an assigned vertex of a target; the shot is valid only if it hits that vertex without passing through any target's interior first. Assign every cow to a unique vertex and minimise the maximum spacing between consecutive cows along the y-axis. Output −1 if no valid assignment exists.

Constraints

Key idea

Each cow's slope and the target geometry determine a small finite set of vertices she can legally hit. Build a bipartite "cow → vertex" feasibility relation, then binary-search on the answer D: a spacing D is achievable iff we can place the 4N cows on the y-axis with no two within D and match each to a valid vertex. The matching feasibility for a fixed sorted y-order reduces to a greedy sweep with a priority queue, because vertices are sortable by y too.

Complexity target

O(N log2 N) — outer binary search on D, inner greedy match with a sorted multiset.

C++ reference solution

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

// Sketch only: full geometry of "shot passes through target interior" is non-trivial.
// Per slope s and vertex v, the segment from (0, y) to v has a closed-form y; we precompute
// per cow the set of legal vertices, then binary-search on D and greedy-match.

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T; cin >> T;
    while (T--) {
        int N; ll X1; cin >> N >> X1;
        vector<array<ll,3>> rects(N);
        for (auto& r : rects) cin >> r[0] >> r[1] >> r[2];
        vector<ll> s(4*N);
        for (auto& x : s) cin >> x;
        // ... build feasibility, binary-search on D, output answer or -1.
        cout << -1 << '\n'; // placeholder
    }
    return 0;
}

This is a placeholder skeleton — the full geometric feasibility check is the hard part of the problem and is best understood by working through the official editorial first.

Pitfalls

Problem 2 · Test Tubes

Official problem (cpid=1399)

Statement (paraphrased)

Two test tubes each hold a column of N units of liquid in colours 1 and 2. A third tube is initially empty. A pour moves the entire top contiguous run of one colour from a source tube to a destination tube (legal only if the destination's top is the same colour or it's empty). Achieve a configuration where tube 1 contains only colour 1 and tube 2 contains only colour 2, using the minimum number of pours. The query type P controls what the judge wants: just the minimum, any near-optimal sequence (≤ M + 5), or an exactly optimal sequence.

Constraints

Key idea

Compress each tube into its maximal runs of equal colour. Each pour removes exactly the top run of the source. The minimum number of pours equals (number of colour-changes in tube 1) + (number of colour-changes in tube 2) + a parity correction that depends on whether the colour at the bottom of each tube matches its target colour. For P=2, you can be lazy and emit ≤ M+5 by using tube 3 as a swap buffer per run. For P=3, you need a precise constructive scheme — typically: shuttle wrong-colour runs through tube 3 in the right order, with one tube playing "storage".

Complexity target

O(N) per test, O(M) output length.

C++ reference solution

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T; cin >> T;
    while (T--) {
        int N, P; cin >> N >> P;
        string a, b; cin >> a >> b;
        // compress each tube into runs
        auto runs = [](const string& s){
            vector<pair<char,int>> r;
            for (char c : s) {
                if (!r.empty() && r.back().first == c) r.back().second++;
                else r.push_back({c, 1});
            }
            return r;
        };
        auto ra = runs(a), rb = runs(b);
        int changesA = (int)ra.size() - 1;
        int changesB = (int)rb.size() - 1;
        int M = changesA + changesB;
        // Parity correction: if the bottom of tube 1 isn't '1' (or analogous for 2), add 1.
        if (!ra.empty() && ra.front().first != '1') M++;
        if (!rb.empty() && rb.front().first != '2') M++;

        if (P == 1) { cout << M << '\n'; continue; }
        // P=2 / P=3: emit a constructive sequence (omitted for brevity).
        cout << M << '\n';
    }
    return 0;
}

Pitfalls

Problem 3 · Moorbles

Official problem (cpid=1400)

Statement (paraphrased)

Elsie starts with N marbles and plays M rounds. In round j Bessie picks one value aj from a K-element menu, and Elsie commits in advance to a guess "Even" or "Odd". A correct guess earns aj marbles; a wrong guess loses aj marbles (capped at her current pile). Elsie wants the lexicographically smallest M-token guess sequence such that, against every possible sequence of Bessie's choices, she never finishes with zero marbles. Output the sequence, or −1.

Constraints

Key idea

For each round j, let mnj = min(aij), mxj = max(aij), and let parities Pj ⊆ {Even, Odd} be the set of parities Bessie can force. If both Even and Odd are forceable, Elsie can't tell what she'll lose — she will lose mxj in the worst case. If only one parity is forceable, she still loses (or gains) deterministically. Reduce to: per round there is a worst-case net delta given Elsie's guess. The lex-min greedy is: scan rounds 1…M, tentatively pick "Even" and verify a future-feasibility test (Elsie's worst-case ending pile is positive) using a suffix sum of "loss-if-Even" vs "loss-if-Odd". If feasible commit Even, else try Odd, else output −1.

Complexity target

O(M) per test after O(MK) preprocessing for the worst-case deltas.

C++ reference solution

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T; cin >> T;
    while (T--) {
        ll N; int M, K; cin >> N >> M >> K;
        vector<vector<int>> a(M, vector<int>(K));
        for (auto& row : a) for (auto& x : row) cin >> x;

        // For each round, compute worst-case delta if Elsie guesses Even vs Odd.
        // Bessie picks the choice that minimises Elsie's pile.
        vector<ll> dEven(M), dOdd(M);
        for (int j = 0; j < M; ++j) {
            ll best_for_even = LLONG_MAX, best_for_odd = LLONG_MAX;
            for (int v : a[j]) {
                ll d_e = (v % 2 == 0) ? +v : -v;
                ll d_o = (v % 2 == 1) ? +v : -v;
                best_for_even = min(best_for_even, d_e);
                best_for_odd  = min(best_for_odd,  d_o);
            }
            dEven[j] = best_for_even;
            dOdd[j]  = best_for_odd;
        }
        // suffix min-pile feasibility: from round j onward, can Elsie keep pile > 0
        // if she greedily picks max(d) each future round?
        vector<ll> bestSuffix(M + 1, 0);
        for (int j = M - 1; j >= 0; --j)
            bestSuffix[j] = bestSuffix[j+1] + max(dEven[j], dOdd[j]);

        ll pile = N;
        vector<string> ans(M);
        bool ok = true;
        for (int j = 0; j < M && ok; ++j) {
            // Try Even first (lex-smaller).
            if (pile + dEven[j] + bestSuffix[j+1] > 0) {
                ans[j] = "Even"; pile += dEven[j];
            } else if (pile + dOdd[j] + bestSuffix[j+1] > 0) {
                ans[j] = "Odd";  pile += dOdd[j];
            } else ok = false;
        }
        if (!ok) cout << -1 << '\n';
        else {
            for (int j = 0; j < M; ++j) cout << ans[j] << " \n"[j == M-1];
        }
    }
    return 0;
}

Pitfalls