引数総和に関する等式制約下の互いに独立な一引数離散凸関数和の最小化
(combinatorial_opt/convex_sum.hpp)

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• Last update: 2022-01-08 20:23:44+09:00
• Include: #include "combinatorial_opt/convex_sum.hpp"
• Link: https://codeforces.com/contest/1344/problem/D
• Link: https://yukicoder.me/problems/no/1495

コスト関数

$\displaystyle y = \sum_{i=1}^N \sum_{j=1}^{k_i} f_i (x_{ij}) $

を，等式制約 $\displaystyle \sum_{i=1}^N \sum_{j=1}^{k_i} x_{ij} = C$ のもと最小化する．

問題例

• Codeforces Round #639 (Div. 1) D. Résumé Review
• yukicoder No.1495 パンの仕入れ

Verified with

• combinatorial_opt/test/convex_sum.test.cpp

Code

#pragma once
#include <cassert>
#include <utility>
#include <vector>

// ax^2 + bx + c (convex), lb <= x <= ub
struct Quadratic {
    using Int = long long;
    Int a, b, c, lb, ub;
    Quadratic(Int a, Int b, Int c, Int lb, Int ub) : a(a), b(b), c(c), lb(lb), ub(ub) {}
    Int slope(Int s) const noexcept {
        if (a == 0) return b <= s ? ub : lb;
        auto ret = (s + a - b) / (a * 2);
        return ret > ub ? ub : ret < lb ? lb : ret;
    }
    Int eval(Int x) const noexcept { return (a * x + b) * x + c; }
    // f(x) - f(x - 1)
    Int next_cost(Int x) const noexcept { return 2 * a * x - a + b; }
};

// x^3 - ax, x \geq 0 (convex)
struct Cubic {
    int a, lb, ub;
    Cubic(int a, int ub) : a(a), lb(0), ub(ub) {}
    int slope(long long s) const noexcept {
        int lo = lb, hi = ub + 1;
        while (hi - lo > 1) {
            int x = (lo + hi) / 2;
            (next_cost(x) <= s ? lo : hi) = x;
        }
        return lo;
    }
    long long eval(long long x) const noexcept { return (x * x - a) * x; }
    // f(x) - f(x - 1)
    long long next_cost(long long x) const noexcept { return 3 * x * x - 3 * x + 1 - a; }
};

// \minimize $\sum_i \sum_j^{k_i} f_i(x_{ij})$
// https://codeforces.com/contest/1344/problem/D
// https://yukicoder.me/problems/no/1495
// return: (y, [[(x_i, # of such x_i), ... ], ...])
template <typename F, typename Int, Int INF>
std::pair<Int, std::vector<std::vector<std::pair<Int, Int>>>>
MinConvexSumUnderLinearConstraint(const std::vector<Int> &k, const std::vector<F> &f, Int C) {
    assert(k.size() == f.size());
    assert(k.size() > 0);
    Int lbsum = 0, ubsum = 0;
    for (auto func : f) lbsum += func.lb, ubsum += func.ub;
    if (lbsum > C or ubsum < C) return {};

    const int N = k.size();
    Int few = -INF, enough = INF;
    while (enough - few > 1) {
        auto slope = few + (enough - few) / 2;
        Int cnt = 0;
        for (int i = 0; i < N; i++) {
            auto tmp = f[i].slope(slope);
            cnt += tmp * k[i];
            if (cnt >= C) break;
        }
        (cnt >= C ? enough : few) = slope;
    }

    std::vector<std::vector<std::pair<Int, Int>>> ret(N);
    std::vector<int> additional;
    Int ctmp = 0;
    Int sol = 0;
    for (int i = 0; i < N; i++) {
        auto xlo = f[i].slope(few);
        auto xhi = f[i].slope(few + 1);
        ctmp += k[i] * xlo;
        ret[i].emplace_back(xlo, k[i]);
        if (xlo < xhi) additional.push_back(i);
        sol += k[i] * f[i].eval(xlo);
    }
    sol += (C - ctmp) * (few + 1);
    while (additional.size()) {
        int i = additional.back();
        additional.pop_back();
        Int add = C - ctmp > k[i] ? k[i] : C - ctmp;
        auto x = ret[i][0].first;
        if (add) {
            ret[i][0].second -= add;
            if (ret[i][0].second == 0) ret[i].pop_back();
            ret[i].emplace_back(x + 1, add);
            ctmp += add;
        }
    }
    return {sol, ret};
}

#line 2 "combinatorial_opt/convex_sum.hpp"
#include <cassert>
#include <utility>
#include <vector>

// ax^2 + bx + c (convex), lb <= x <= ub
struct Quadratic {
    using Int = long long;
    Int a, b, c, lb, ub;
    Quadratic(Int a, Int b, Int c, Int lb, Int ub) : a(a), b(b), c(c), lb(lb), ub(ub) {}
    Int slope(Int s) const noexcept {
        if (a == 0) return b <= s ? ub : lb;
        auto ret = (s + a - b) / (a * 2);
        return ret > ub ? ub : ret < lb ? lb : ret;
    }
    Int eval(Int x) const noexcept { return (a * x + b) * x + c; }
    // f(x) - f(x - 1)
    Int next_cost(Int x) const noexcept { return 2 * a * x - a + b; }
};

// x^3 - ax, x \geq 0 (convex)
struct Cubic {
    int a, lb, ub;
    Cubic(int a, int ub) : a(a), lb(0), ub(ub) {}
    int slope(long long s) const noexcept {
        int lo = lb, hi = ub + 1;
        while (hi - lo > 1) {
            int x = (lo + hi) / 2;
            (next_cost(x) <= s ? lo : hi) = x;
        }
        return lo;
    }
    long long eval(long long x) const noexcept { return (x * x - a) * x; }
    // f(x) - f(x - 1)
    long long next_cost(long long x) const noexcept { return 3 * x * x - 3 * x + 1 - a; }
};

// \minimize $\sum_i \sum_j^{k_i} f_i(x_{ij})$
// https://codeforces.com/contest/1344/problem/D
// https://yukicoder.me/problems/no/1495
// return: (y, [[(x_i, # of such x_i), ... ], ...])
template <typename F, typename Int, Int INF>
std::pair<Int, std::vector<std::vector<std::pair<Int, Int>>>>
MinConvexSumUnderLinearConstraint(const std::vector<Int> &k, const std::vector<F> &f, Int C) {
    assert(k.size() == f.size());
    assert(k.size() > 0);
    Int lbsum = 0, ubsum = 0;
    for (auto func : f) lbsum += func.lb, ubsum += func.ub;
    if (lbsum > C or ubsum < C) return {};

    const int N = k.size();
    Int few = -INF, enough = INF;
    while (enough - few > 1) {
        auto slope = few + (enough - few) / 2;
        Int cnt = 0;
        for (int i = 0; i < N; i++) {
            auto tmp = f[i].slope(slope);
            cnt += tmp * k[i];
            if (cnt >= C) break;
        }
        (cnt >= C ? enough : few) = slope;
    }

    std::vector<std::vector<std::pair<Int, Int>>> ret(N);
    std::vector<int> additional;
    Int ctmp = 0;
    Int sol = 0;
    for (int i = 0; i < N; i++) {
        auto xlo = f[i].slope(few);
        auto xhi = f[i].slope(few + 1);
        ctmp += k[i] * xlo;
        ret[i].emplace_back(xlo, k[i]);
        if (xlo < xhi) additional.push_back(i);
        sol += k[i] * f[i].eval(xlo);
    }
    sol += (C - ctmp) * (few + 1);
    while (additional.size()) {
        int i = additional.back();
        additional.pop_back();
        Int add = C - ctmp > k[i] ? k[i] : C - ctmp;
        auto x = ret[i][0].first;
        if (add) {
            ret[i][0].second -= add;
            if (ret[i][0].second == 0) ret[i].pop_back();
            ret[i].emplace_back(x + 1, add);
            ctmp += add;
        }
    }
    return {sol, ret};
}

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