题解 | #魔法相册的重复记忆#

#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <sstream>
#include <algorithm>

using namespace std;

// 定义结构体存储照片信息
struct Photo {
    int id;
    int t;
    int count;
};

// 排序规则：按时间戳升序
bool compare(const Photo& a, const Photo& b) {
    return a.t < b.t;
}

int main() {
    int n;
    // 读取相册数量
    cin >> n;
    string dummy;
    getline(cin, dummy); // 消耗掉第一行后的换行符

    map<int, int> counts;     // 存储 ID 出现的次数
    map<int, int> timestamps; // 存储 ID 对应的时间戳

    for (int i = 0; i < n; ++i) {
        string line;
        getline(cin, line);
        stringstream ss(line);
        int id, t;
        // 每两个整数为一组 (ID, T)
        while (ss >> id >> t) {
            counts[id]++;
            timestamps[id] = t;
        }
    }

    vector<Photo> result;
    for (auto const& [id, count] : counts) {
        if (count > 1) {
            result.push_back({id, timestamps[id], count});
        }
    }

    // 排序
    sort(result.begin(), result.end(), compare);

    // 格式化输出
    for (int i = 0; i < result.size(); ++i) {
        cout << result[i].id << " " << result[i].count << (i == result.size() - 1 ? "" : " ");
    }
    cout << endl;

    return 0;
}

import java.util.*;

public class Main {
    // 存储照片信息的类
    static class Photo {
        int id;
        int t;
        int count;
        Photo(int id, int t, int count) {
            this.id = id;
            this.t = t;
            this.count = count;
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        // 读取相册数量
        int n = Integer.parseInt(sc.nextLine());

        Map<Integer, Integer> counts = new HashMap<>();     // ID 频次统计
        Map<Integer, Integer> timestamps = new HashMap<>(); // ID 时间戳映射

        for (int i = 0; i < n; i++) {
            String line = sc.nextLine();
            String[] parts = line.trim().split("\\s+");
            for (int j = 0; j < parts.length; j += 2) {
                int id = Integer.parseInt(parts[j]);
                int t = Integer.parseInt(parts[j + 1]);
                counts.put(id, counts.getOrDefault(id, 0) + 1);
                timestamps.put(id, t);
            }
        }

        List<Photo> result = new ArrayList<>();
        for (Map.Entry<Integer, Integer> entry : counts.entrySet()) {
            if (entry.getValue() > 1) {
                int id = entry.getKey();
                result.add(new Photo(id, timestamps.get(id), entry.getValue()));
            }
        }

        // 按时间戳排序
        result.sort((a, b) -> a.t - b.t);

        // 输出结果
        for (int i = 0; i < result.size(); i++) {
            System.out.print(result.get(i).id + " " + result.get(i).count + (i == result.size() - 1 ? "" : " "));
        }
        System.out.println();
    }
}

def solve():
    # 读取相册数量
    n = int(input())
    counts = {}      # ID 频次统计
    timestamps = {}  # ID 时间戳映射

    for _ in range(n):
        line_data = input().split()
        # 每两个整数为一组 (ID, T)
        for i in range(0, len(line_data), 2):
            photo_id = int(line_data[i])
            t = int(line_data[i + 1])
            counts[photo_id] = counts.get(photo_id, 0) + 1
            timestamps[photo_id] = t

    res_list = []
    for photo_id, count in counts.items():
        if count > 1:
            res_list.append((photo_id, count, timestamps[photo_id]))

    # 按时间戳 T 升序排序
    res_list.sort(key=lambda x: x[2])

    output = []
    for photo_id, count, t in res_list:
        output.append(str(photo_id))
        output.append(str(count))
    
    print(" ".join(output))

solve()