CCF CAT- 全国算法精英大赛（2024第二场）往届真题练习 4 | 珂学家

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前言

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餐馆

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v
        int 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dp2
        dp2[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bz
            int 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
        }
    }

    static
    class 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.readline

n, 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] = 1

def 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)))

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