思路：可撤销的0-1背包

考察了多个知识点，包括

• 差分技巧
• 离线思路
• 0-1背包

不过这题卡语言，尤其卡python

import java.io.*;
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;public class Main {static final long mod = (long)1e9 + 7;public static void main(String[] args) {AReader scanner = new AReader();PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));// 读取n和vint n = scanner.nextInt();int v = scanner.nextInt();// 读取并处理物品信息List<int[]> packs = new ArrayList<>();List<int[]> ops = new ArrayList<>();for (int i = 0; i < n; i++) {int s = scanner.nextInt();int e = scanner.nextInt();int w = scanner.nextInt();packs.add(new int[]{s, e, w});ops.add(new int[]{s, 1, w});ops.add(new int[]{e + 1, -1, w});}// 对ops按时间排序ops.sort(Comparator.comparingInt(a -> a[0]));// 读取q和查询时间int q = scanner.nextInt();int[] arr = IntStream.range(0, q).map(i -> scanner.nextInt()).toArray();// 将查询和索引关联起来List<int[]> qs = IntStream.range(0, q).mapToObj(i -> new int[]{arr[i], i}).collect(Collectors.toList());qs.sort(Comparator.comparingInt(a -> a[0]));// 互斥的两类long[] dp1 = new long[v + 1];long[] dp2 = new long[v + 1];// 初始化dp2dp2[0] = 1;for (int[] pack : packs) {int s = pack[0];int w = pack[2];for (int m = v - w; m >= 0; m--) {dp2[m + w] += dp2[m];dp2[m + w] %= mod;}}dp1[0] = 1;// 双指针，离散做法int p1 = 0;int p2 = 0;int[][] res = new int[q][2];for (int i = 0; i < 101; i++) { // 假设结束时间不超过arr[q-1]while (p1 < ops.size() && ops.get(p1)[0] <= i) {int[] op = ops.get(p1);int d = op[1];int w = op[2];if (d == 1) {add01(dp1, w, v);remove01(dp2, w, v);} else {remove01(dp1, w, v);add01(dp2, w, v);}p1++;}// 找到fz和bzint fz = 0;for (int j = v; j >= 0; j--) {if (dp1[j] > 0) {fz = j;break;}}int bz = 0;for (int j = v - fz; j >= 0; j--) {if (dp2[j] > 0) {bz = j;break;}}// 填充结果while (p2 < qs.size() && qs.get(p2)[0] <= i) {res[qs.get(p2)[1]][0] = fz;res[qs.get(p2)[1]][1] = bz;p2++;}}// 输出结果for (int[] pair : res) {out.println(pair[0] + " " + pair[1]);}out.flush();out.close();}private static void add01(long[] dp, int w, int v) {for (int u = v - w; u >= 0; u--) {dp[u + w] += dp[u];dp[u + w] %= mod;}}private static void remove01(long[] dp, int w, int v) {for (int u = 0; u <= v - w; u++) {dp[u + w] -= dp[u];dp[u + w] = (dp[u + w] % mod + mod) % mod;// 注意：在Java中，如果dp[u]是0或负数，可能需要额外的逻辑来避免负数}}staticclass AReader {private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));private StringTokenizer tokenizer = new StringTokenizer("");private String innerNextLine() {try {return reader.readLine();} catch (IOException ex) {return null;}}public boolean hasNext() {while (!tokenizer.hasMoreTokens()) {String nextLine = innerNextLine();if (nextLine == null) {return false;}tokenizer = new StringTokenizer(nextLine);}return true;}public String nextLine() {tokenizer = new StringTokenizer("");return innerNextLine();}public String next() {hasNext();return tokenizer.nextToken();}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}}}

python 版本被卡常

# coding=utf-8
# coding=utf-8
import sys
input=sys.stdin.buffer.readlinen, v = list(map(int, input().split()))ops = []
packs = []
for i in range(n):s, e, w = list(map(int, input().split()))packs.append((s, e, w))ops.append((s, 1, w))ops.append((e + 1, -1, w))
ops.sort(key=lambda x: x[0])t1, t2 = 100, 0
q = int(input())
arr = list(map(int, input().split()))
qs = []
for i in range(q):qs.append((arr[i], i))t1 = min(t1, arr[i])t2 = max(t2, arr[i])
qs.sort(key=lambda x: [0])# 互斥的两类
dp1 = [0] * (v + 1)
dp2 = [0] * (v + 1)#--------------------------------dp2[0] = 1
for (s, e, w) in packs:for m in range(v - w, -1, -1):dp2[m + w] += dp2[m]dp1[0] = 1def add01(dp, w):for u in range(v - w, -1, -1):dp[u + w] += dp[u]def remove01(dp, w):for u in range(0, v - w + 1):dp[u + w] -= dp[u]# 双指针，离散做法
res = [[]] * q
p1, p2 = 0, 0
for i in range(t1, t2 + 1):while p1 < len(ops) and ops[p1][0] <= i:d = ops[p1][1]if d == 1:add01(dp1, ops[p1][2])remove01(dp2, ops[p1][2])elif d == -1:remove01(dp1, ops[p1][2])add01(dp2, ops[p1][2])p1 += 1# print (i, dp1, dp2)fz = max([i for i in range(v + 1) if dp1[i] > 0])bz = max([i for i in range(v + 1) if dp2[i] > 0 and i <= (v - fz)])while p2 < len(qs) and qs[p2][0] <= i:res[qs[p2][1]] = (fz, bz)p2 +=  1# for (fz, bz) in res:
#     print (fz, bz)print ("\n".join(map(lambda x: str(x[0]) + " "+ str(x[1]), res)))