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#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(w, m, v, W)
: 価値の最大値を求める.#include "knapsack-limitations.hpp"
template <typename T>
T knapsack_limitations(const vector<T> &w, const vector<T> &m,
const vector<int> &v, const T &W) {
const int N = (int)w.size();
auto v_max = *max_element(begin(v), end(v));
if (v_max == 0) return 0;
vector<int> ma(N);
vector<T> mb(N);
for (int i = 0; i < N; i++) {
ma[i] = 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), less<>());
vector<int> ord(N);
iota(begin(ord), end(ord), 0);
sort(begin(ord), end(ord),
[&](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 = min(mb[p], rest / w[p]);
if (get <= 0) continue;
cost += get * v[p];
rest -= get * w[p];
}
ret = max(ret, cost);
}
return ret;
}
#line 1 "dp/knapsack-limitations.hpp"
template <typename T, typename Compare = greater<T> >
vector<T> knapsack_limitations(const vector<int> &w, const vector<int> &m,
const vector<T> &v, const int &W, const T &NG,
const Compare &comp = Compare()) {
const int N = (int)w.size();
vector<T> dp(W + 1, NG), deqv(W + 1);
dp[0] = T();
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 2 "dp/knapsack-limitations-2.hpp"
template <typename T>
T knapsack_limitations(const vector<T> &w, const vector<T> &m,
const vector<int> &v, const T &W) {
const int N = (int)w.size();
auto v_max = *max_element(begin(v), end(v));
if (v_max == 0) return 0;
vector<int> ma(N);
vector<T> mb(N);
for (int i = 0; i < N; i++) {
ma[i] = 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), less<>());
vector<int> ord(N);
iota(begin(ord), end(ord), 0);
sort(begin(ord), end(ord),
[&](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 = min(mb[p], rest / w[p]);
if (get <= 0) continue;
cost += get * v[p];
rest -= get * w[p];
}
ret = max(ret, cost);
}
return ret;
}