結果

問題 No.2026 Yet Another Knapsack Problem
ユーザー suisen
提出日時 2022-04-22 14:36:08
言語 Java
(openjdk 23)
結果
AC  
実行時間 3,183 ms / 10,000 ms
コード長 2,610 bytes
コンパイル時間 2,199 ms
コンパイル使用メモリ 79,784 KB
実行使用メモリ 129,736 KB
最終ジャッジ日時 2024-07-19 12:47:56
合計ジャッジ時間 31,634 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 42
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 #
プレゼンテーションモードにする

import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
var sc = new Scanner(System.in);
var pw = new PrintWriter(System.out, false);
solve(sc, pw);
sc.close();
pw.flush();
pw.close();
}
public static final long INF = 1L << 60;
public static void solve(Scanner sc, PrintWriter pw) {
final int n = Integer.parseInt(sc.next());
var c = new int[n + 1];
var v = new long[n + 1];
for (int i = 1; i <= n; i++) {
c[i] = Integer.parseInt(sc.next());
v[i] = Integer.parseInt(sc.next());
}
var dp = new long[n + 1][n + 1];
for (int i = 0; i <= n; i++) Arrays.fill(dp[i], -INF);
dp[0][0] = 0;
for (int i = n; i > 0; i--) {
final int max_num = n / i;
final int max_sum = n;
for (int num = 0; num <= max_num; num++) {
final int sum_r = num > 0 ? i - 1 : max_sum;
for (int sum = 0; sum <= sum_r; sum++) {
final int max_p = Math.min(max_num - num, (max_sum - sum) / i);
var init = new long[max_p + 1];
for (int p = 0; p <= max_p; ++p) {
init[p] = dp[num + p][sum + i * p] - (num + p) * v[i];
}
var seg = new RangeMaxSegmentTree(init);
for (int p = 0; p <= max_p; ++p) {
dp[num + p][sum + i * p] = seg.max(Math.max(0, p - c[i]), p + 1) + (num + p) * v[i];
}
}
}
}
for (int i = 1; i <= n; i++) {
pw.println(Arrays.stream(dp[i]).max().getAsLong());
}
}
public static class RangeMaxSegmentTree {
private int _n;
private long[] _dat;
public RangeMaxSegmentTree(long[] dat) {
_n = dat.length;
_dat = new long[2 * _n];
System.arraycopy(dat, 0, _dat, _n, _n);
for (int i = _n - 1; i > 0; i--) {
_dat[i] = Math.max(_dat[2 * i], _dat[2 * i + 1]);
}
}
public long max(int l, int r) {
long res = Long.MIN_VALUE;
l += _n;
r += _n;
while (l < r) {
if ((l & 1) != 0) res = Math.max(res, _dat[l++]);
if ((r & 1) != 0) res = Math.max(res, _dat[--r]);
l >>= 1;
r >>= 1;
}
return res;
}
}
}
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