結果

問題 No.1316 Maximum Minimum Spanning Tree
コンテスト
ユーザー hitonanode
提出日時 2020-12-12 00:21:57
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 38 ms / 2,000 ms
コード長 5,391 bytes
コンパイル時間 1,913 ms
コンパイル使用メモリ 117,976 KB
最終ジャッジ日時 2025-01-16 22:28:52
ジャッジサーバーID
（参考情報） 		judge3 / judge4
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 4
other AC * 78
権限があれば一括ダウンロードができます

ソースコード

diff #
raw source code 

#include <algorithm>
#include <functional>
#include <iostream>
#include <limits>
#include <numeric>
#include <queue>
#include <utility>
#include <vector>


// MaxFlow (Dinic algorithm)
template <typename T> struct MaxFlow {
    struct edge {
        int to;
        T cap;
        int rev;
    };
    std::vector<std::vector<edge>> edges;
    std::vector<int> level; // level[i] = distance between vertex S and i (Default: -1)
    std::vector<int> iter;  // iteration counter, used for Dinic's DFS

    void bfs(int s) {
        level.assign(edges.size(), -1);
        std::queue<int> q;
        level[s] = 0;
        q.push(s);
        while (!q.empty()) {
            int v = q.front();
            q.pop();
            for (edge &e : edges[v]) {
                if (e.cap > 0 and level[e.to] < 0) {
                    level[e.to] = level[v] + 1;
                    q.push(e.to);
                }
            }
        }
    }

    T dfs_dinic(int v, int goal, T f) {
        if (v == goal) return f;
        for (int &i = iter[v]; i < (int)edges[v].size(); i++) {
            edge &e = edges[v][i];
            if (e.cap > 0 and level[v] < level[e.to]) {
                T d = dfs_dinic(e.to, goal, std::min(f, e.cap));
                if (d > 0) {
                    e.cap -= d;
                    edges[e.to][e.rev].cap += d;
                    return d;
                }
            }
        }
        return 0;
    }

    MaxFlow(int N) { edges.resize(N); }
    void add_edge(int from, int to, T capacity) {
        edges[from].push_back(edge{to, capacity, (int)edges[to].size()});
        edges[to].push_back(edge{from, (T)0, (int)edges[from].size() - 1});
    }
    // Dinic algorithm
    // Complexity: O(V^2 E)
    T Dinic(int s, int t, T req) {
        T flow = 0;
        while (req > 0) {
            bfs(s);
            if (level[t] < 0) break;
            iter.assign(edges.size(), 0);
            T f;
            while ((f = dfs_dinic(s, t, req)) > 0) flow += f, req -= f;
        }
        return flow;
    }
    T Dinic(int s, int t) { return Dinic(s, t, std::numeric_limits<T>::max()); }
};


// LinearProgrammingOnBasePolyhedron : Maximize/minimize linear function on base polyhedron, using Edmonds' algorithm
//
// maximize/minimize cx s.t. (x on some base polyhedron)
// Reference: <https://www.amazon.co.jp/dp/B01N6G0579>, Sec. 2.4, Algorithm 2.2-2.3
//            "Submodular Functions, Matroids, and Certain Polyhedra" [Edmonds+, 1970]
template <typename Tvalue> struct LinearProgrammingOnBasePolyhedron {
    using Tfunc = std::function<Tvalue(int, const std::vector<Tvalue> &)>;
    static Tvalue EPS;
    int N;
    std::vector<Tvalue> c;
    Tfunc maximize_xi;
    Tvalue xsum;
    bool minimize;

    Tvalue fun;
    std::vector<Tvalue> x;
    bool infeasible;

    void _init(const std::vector<Tvalue> &c_, Tfunc q_, Tvalue xsum_, Tvalue xlowerlimit, bool minimize_) {
        N = c_.size();
        c = c_;
        maximize_xi = q_;
        xsum = xsum_;
        minimize = minimize_;
        fun = 0;
        x.assign(N, xlowerlimit);
        infeasible = false;
    }

    void _solve() {
        std::vector<std::pair<Tvalue, int>> c2i(N);
        for (int i = 0; i < N; i++) c2i[i] = std::make_pair(c[i], i);

        std::sort(c2i.begin(), c2i.end());
        if (!minimize) std::reverse(c2i.begin(), c2i.end());
        for (const auto &p : c2i) {
            const int i = p.second;
            x[i] = maximize_xi(i, x);
        }
        if (std::abs(std::accumulate(x.begin(), x.end(), Tvalue(0)) - xsum) > EPS) {
            infeasible = true;
        } else {
            for (int i = 0; i < N; i++) fun += x[i] * c[i];
        }
    }

    LinearProgrammingOnBasePolyhedron(const std::vector<Tvalue> &c_, Tfunc q_, Tvalue xsum_, Tvalue xlowerlimit, bool minimize_) {
        _init(c_, q_, xsum_, xlowerlimit, minimize_);
        _solve();
    }
};
template <> long long LinearProgrammingOnBasePolyhedron<long long>::EPS = 0;
template <> long double LinearProgrammingOnBasePolyhedron<long double>::EPS = 1e-12;

using std::cin, std::cout, std::vector;

int main() {
    using Num = long long;
    int N, M;
    long long K;
    cin >> N >> M >> K;
    vector<int> A(M), B(M);
    vector<Num> C(M), D(M);
    for (int i = 0; i < M; i++) {
        cin >> A[i] >> B[i] >> C[i] >> D[i];
        A[i]--, B[i]--;
    }

    auto maximize_xi = [&](int ie, const vector<Num> &xnow) -> Num {
        MaxFlow<Num> mf(N + 2);
        mf.add_edge(N, A[ie], 2 * K * N);
        mf.add_edge(N, B[ie], 2 * K * N);
        for (int je = 0; je < M; je++) {
            mf.add_edge(A[je], B[je], xnow[je]);
            mf.add_edge(B[je], A[je], xnow[je]);
            mf.add_edge(N, A[je], xnow[je]);
            mf.add_edge(N, B[je], xnow[je]);
        }
        for (int iv = 0; iv < N; iv++) mf.add_edge(iv, N + 1, 2 * K);
        Num ret = mf.Dinic(N, N + 1, 1LL << 60) / 2 - K - std::accumulate(xnow.begin(), xnow.end(), (Num)0);
        return std::min(ret, D[ie]);
    };

    LinearProgrammingOnBasePolyhedron<Num> solver(C, maximize_xi, K * (N - 1), 0, true);
    if (solver.infeasible) {
        cout << "-1\n";
    } else {
        cout << (long long)solver.fun << '\n';
    }
}
0