#1166617 (C++17) No.3585 Make Ends Meet (Easy)

• 問題: No.3585 Make Ends Meet (Easy)
• ユーザー: marc2825
• 提出日時: 2026-05-25 09:21:42
• 言語: C++17 (gcc 15.2.0 + boost 1.90.0)
• 結果: AC
• 実行時間: 2 ms / 2,000 ms
• コード長: 2,888 bytes
• コンパイル時間: 1,282 ms
• コンパイル使用メモリ: 222,304 KB
• 実行使用メモリ: 5,888 KB
• 最終ジャッジ日時: 2026-07-10 20:51:32
• 合計ジャッジ時間: 3,337 ms
• ジャッジサーバーID （参考情報）: judge3_0 / judge1_0

ファイルパターン	結果
sample	AC * 2
other	AC * 48

ソースコード

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

int N, M, K;
vector<pair<int, int>> ans;

void print_no() {
    cout << "No\n";
}

void print_yes() {
    cout << "Yes\n";
    for (auto [u, v] : ans) {
        cout << u << ' ' << v << '\n';
    }
}

int main() {
    cin >> N >> M >> K;

    int T = N * (N - 1) / 2;

    // WRONG:
    // This solution forgets the special case K == 1.
    // The following formula and construction are intended for K >= 2.
    int min_removed = K * (K - 1) / 2 + (N - K - 1) * (K - 2);
    int max_removed = T - K;

    if (M < min_removed || M > max_removed) {
        print_no();
        return 0;
    }

    // Use the path 1 - 2 - ... - K - N.
    // For K == 1, this becomes 1 - N, but the layer construction below
    // is not valid for the intended K >= 2 reasoning.
    vector<int> path;
    path.push_back(1);
    for (int v = 2; v <= K; ++v) {
        path.push_back(v);
    }
    path.push_back(N);

    vector<vector<bool>> is_path_edge(N + 1, vector<bool>(N + 1, false));
    for (int i = 0; i + 1 < (int)path.size(); ++i) {
        int a = path[i];
        int b = path[i + 1];
        if (a > b) swap(a, b);
        is_path_edge[a][b] = true;
    }

    // WRONG for K == 1:
    // layer[1] is first set to 0, but then layer[N] = K = 1.
    // Extra vertices 2..N-1 are also put into layer 1.
    // This construction does not represent "distance exactly 1" correctly.
    vector<int> layer(N + 1, -1);
    layer[1] = 0;
    for (int i = 1; i <= K - 1; ++i) {
        layer[i + 1] = i;
    }
    layer[N] = K;

    // Make extra vertices belong to layer 1.
    for (int v = K + 1; v <= N - 1; ++v) {
        layer[v] = 1;
    }

    vector<vector<bool>> removed(N + 1, vector<bool>(N + 1, false));

    // First remove all edges whose layer difference is at least 2.
    // This is the intended minimum-removal construction for K >= 2.
    // For K == 1, no such edges exist, so this does not capture the true condition.
    for (int u = 1; u <= N; ++u) {
        for (int v = u + 1; v <= N; ++v) {
            if (abs(layer[u] - layer[v]) >= 2) {
                removed[u][v] = true;
                ans.push_back({u, v});
            }
        }
    }

    // Remove additional non-path edges arbitrarily until exactly M edges are removed.
    // Because (1, N) is treated as the only path edge when K == 1,
    // this code may still output seemingly plausible answers,
    // but the feasibility check above is wrong for K == 1.
    for (int u = 1; u <= N && (int)ans.size() < M; ++u) {
        for (int v = u + 1; v <= N && (int)ans.size() < M; ++v) {
            if (removed[u][v]) continue;
            if (is_path_edge[u][v]) continue;

            removed[u][v] = true;
            ans.push_back({u, v});
        }
    }

    if ((int)ans.size() < M) {
        print_no();
        return 0;
    }

    print_yes();
    return 0;
}