結果
問題 | No.2026 Yet Another Knapsack Problem |
ユーザー |
|
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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 42 |
ソースコード
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;}}}