Knapsack Limitations (個数制限つきナップサック問題) $O(N^2 \max(v_i)^2)$ (dp/knapsack-limitations-2.hpp)

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• Last update: 2026-08-10 06:54:42+00:00
• Include: #include "dp/knapsack-limitations-2.hpp"

個数制限つきナップサック問題を次に示す。

重さ $w_i$、価値 $v_i$ であるような $N$ 種類の品物がある。$i$ 番目の品物は $m_i$ 個まで選ぶことができる。重さの和が $W$ 以下となるように選ぶとき、価値の最大値を求めよ。

まず「dp[価値の和]:= 重さの和の最小値」をある程度の大きさ($N \max(v_i)^2$)まで求める。残りの分は、効率が良い順(価値/重さが大きい順)に貪欲にとってもいいことが示せて、なんかできる(c)。

knapsack_limitations

T knapsack_limitations(const vector<T>& w, const vector<T>& m, const vector<int>& v, const T& W)

価値の最大値を求める。

計算量

• $O(N^2 \max(v_i)^2)$

Depends on

• Knapsack Limitations (個数制限つきナップサック問題) $O(NW)$ (dp/knapsack-limitations.hpp)

Verified with

• test/verify/aoj-dpl-1-i.test.cpp

Code

#pragma once

#include <algorithm>
#include <functional>
#include <numeric>
#include <vector>

#include "knapsack-limitations.hpp"

template <typename T>
T knapsack_limitations(const std::vector<T>& w, const std::vector<T>& m,
                       const std::vector<int>& v, const T& W) {
  const int N = (int)w.size();
  auto v_max = *std::max_element(v.begin(), v.end());
  if (v_max == 0) return 0;
  std::vector<int> ma(N);
  std::vector<T> mb(N);
  for (int i = 0; i < N; i++) {
    ma[i] = std::min<T>(m[i], v_max - 1);
    mb[i] = m[i] - ma[i];
  }
  T sum = 0;
  for (int i = 0; i < N; i++) sum += ma[i] * v[i];
  auto dp = knapsack_limitations(v, ma, w, sum, T(-1), std::less<>());
  std::vector<int> ord(N);
  std::iota(ord.begin(), ord.end(), 0);
  std::sort(ord.begin(), ord.end(),
            [&](int a, int b) { return v[a] * w[b] > v[b] * w[a]; });
  T ret = T();
  for (int i = 0; i < dp.size(); i++) {
    if (dp[i] > W || dp[i] == -1) continue;
    T rest = W - dp[i], cost = i;
    for (auto& p : ord) {
      auto get = std::min(mb[p], rest / w[p]);
      if (get <= 0) continue;
      cost += get * v[p];
      rest -= get * w[p];
    }
    ret = std::max(ret, cost);
  }
  return ret;
}

#line 2 "dp/knapsack-limitations-2.hpp"

#include <algorithm>
#include <functional>
#include <numeric>
#include <vector>

#line 2 "dp/knapsack-limitations.hpp"

#line 5 "dp/knapsack-limitations.hpp"

template <typename T, typename Compare = std::greater<T> >
std::vector<T> knapsack_limitations(const std::vector<int>& w,
                                    const std::vector<int>& m,
                                    const std::vector<T>& v, const int& W,
                                    const T& NG,
                                    const Compare& comp = Compare()) {
  const int N = (int)w.size();
  std::vector<T> dp(W + 1, NG), deqv(W + 1);
  dp[0] = T();
  std::vector<int> deq(W + 1);
  for (int i = 0; i < N; i++) {
    if (w[i] == 0) {
      for (int j = 0; j <= W; j++) {
        if (dp[j] != NG && comp(dp[j] + v[i] * m[i], dp[j])) {
          dp[j] = dp[j] + v[i] * m[i];
        }
      }
    } else {
      for (int a = 0; a < w[i]; a++) {
        int s = 0, t = 0;
        for (int j = 0; w[i] * j + a <= W; j++) {
          if (dp[w[i] * j + a] != NG) {
            auto val = dp[w[i] * j + a] - j * v[i];
            while (s < t && comp(val, deqv[t - 1])) --t;
            deq[t] = j;
            deqv[t++] = val;
          }
          if (s < t) {
            dp[j * w[i] + a] = deqv[s] + j * v[i];
            if (deq[s] == j - m[i]) ++s;
          }
        }
      }
    }
  }
  return dp;
}
#line 9 "dp/knapsack-limitations-2.hpp"

template <typename T>
T knapsack_limitations(const std::vector<T>& w, const std::vector<T>& m,
                       const std::vector<int>& v, const T& W) {
  const int N = (int)w.size();
  auto v_max = *std::max_element(v.begin(), v.end());
  if (v_max == 0) return 0;
  std::vector<int> ma(N);
  std::vector<T> mb(N);
  for (int i = 0; i < N; i++) {
    ma[i] = std::min<T>(m[i], v_max - 1);
    mb[i] = m[i] - ma[i];
  }
  T sum = 0;
  for (int i = 0; i < N; i++) sum += ma[i] * v[i];
  auto dp = knapsack_limitations(v, ma, w, sum, T(-1), std::less<>());
  std::vector<int> ord(N);
  std::iota(ord.begin(), ord.end(), 0);
  std::sort(ord.begin(), ord.end(),
            [&](int a, int b) { return v[a] * w[b] > v[b] * w[a]; });
  T ret = T();
  for (int i = 0; i < dp.size(); i++) {
    if (dp[i] > W || dp[i] == -1) continue;
    T rest = W - dp[i], cost = i;
    for (auto& p : ord) {
      auto get = std::min(mb[p], rest / w[p]);
      if (get <= 0) continue;
      cost += get * v[p];
      rest -= get * w[p];
    }
    ret = std::max(ret, cost);
  }
  return ret;
}

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