假设全校有最多40000名学生和最多2500门课程。现给出每个学生的选课清单,要求输出每门课的选课学生名单。
输入格式:
输入的第一行是两个正整数:N(≤40000),为全校学生总数;K(≤2500),为总课程数。此后N行,每行包括一个学生姓名(3个大写英文字母+1位数字)、一个正整数C(≤20)代表该生所选的课程门数、随后是C个课程编号。简单起见,课程从1到K编号。
输出格式:
顺序输出课程1到K的选课学生名单。格式为:对每一门课,首先在一行中输出课程编号和选课学生总数(之间用空格分隔),之后在第二行按字典序输出学生名单,每个学生名字占一行。
输入样例:
10 5
ZOE1 2 4 5
ANN0 3 5 2 1
BOB5 5 3 4 2 1 5
JOE4 1 2
JAY9 4 1 2 5 4
FRA8 3 4 2 5
DON2 2 4 5
AMY7 1 5
KAT3 3 5 4 2
LOR6 4 2 4 1 5
输出样例:
1 4
ANN0
BOB5
JAY9
LOR6
2 7
ANN0
BOB5
FRA8
JAY9
JOE4
KAT3
LOR6
3 1
BOB5
4 7
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
5 9
AMY7
ANN0
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
加上了详细的注释,方便题主理解学习,望采纳
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
// 学生人数
int n = scan.nextInt();
// 课程数量
int k = scan.nextInt();
// 记录课程和学生列表的映射
Map<Integer, List<String>> course = new HashMap<>();
// 初始化课程列表
for (int i = 0; i < k; i++) {
course.put(i + 1, new ArrayList<>());
}
// 处理每个学生的课程选择
for (int i = 0; i < n; i++) {
// 学生名字
String name = scan.next();
// 选择课程数量
int num = scan.nextInt();
for (int j = 0; j < num; j++) {
// 选择的课程编号
int courseId = scan.nextInt();
// 把学生加入课程学生列表中
course.get(courseId).add(name);
}
}
// 输出每门课的选课学生名单
for (int i = 1; i <= k; i++) {
// 获取课程学生列表
List<String> students = course.get(i);
// 输出课程编号和选课学生总数
System.out.println(i + " " + students.size());
// 学生名字排序
Collections.sort(students);
// 输出学生名字列表
for (String student : students) {
System.out.println(student);
}
}
}
}
望采纳,代码示例:
import java.util.*;
public class Main {
static int n, k;
static List<String>[] courses = new ArrayList[2510];
static boolean[][] stu = new boolean[2510][40000];
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
k = sc.nextInt();
for (int i = 0; i < 2510; i++) {
courses[i] = new ArrayList<>();
}
for (int i = 0; i < n; i++) {
String name = sc.next();
int c = sc.nextInt();
for (int j = 0; j < c; j++) {
int course = sc.nextInt();
stu[course][i] = true;
}
}
for (int i = 1; i <= k; i++) {
int num = 0;
for (int j = 0; j < n; j++) {
if (stu[i][j]) {
num++;
courses[i].add(getName(j));
}
}
Collections.sort(courses[i]);
System.out.println(i + " " + num);
for (String name : courses[i]) {
System.out.println(name);
}
}
}
private static String getName(int index) {
return String.format("%c%c%c%d", 'A' + (index / 26 / 26), 'A' + index / 26 % 26, 'A' + index % 26, index % 10);
}
}
实现步骤如下:
读入输入数据,存储到学生对象中,包括姓名和选课课程编号。
对每门课程,遍历每个学生,记录选课学生的姓名。
按字典序输出学生名单。
输出每门课程的选课学生列表。
直接上代码,望采纳
import java.util.*;
import java.io.*;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int k = Integer.parseInt(s[1]);
Map<Integer, List<String>> map = new TreeMap<>();
for (int i = 1; i <= k; i++) {
map.put(i, new ArrayList<>());
}
for (int i = 0; i < n; i++) {
s = br.readLine().split(" ");
int m = Integer.parseInt(s[1]);
for (int j = 0; j < m; j++) {
int course = Integer.parseInt(s[j + 2]);
map.get(course).add(s[0]);
}
}
for (int i = 1; i <= k; i++) {
List<String> list = map.get(i);
System.out.println(i + " " + list.size());
Collections.sort(list);
for (String name : list) {
System.out.println(name);
}
}
}
}
思路很清晰,可以直接写代码,
如果在写代码过程中有疑问,可以来交流。
如下是使用 Java 解决这道题的代码:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
Map<Integer, List<String>> courseMap = new HashMap<>();
Map<String, List<Integer>> studentMap = new HashMap<>();
for (int i = 0; i < n; i++) {
String name = sc.next();
int c = sc.nextInt();
List<Integer> courses = new ArrayList<>();
for (int j = 0; j < c; j++) {
int courseId = sc.nextInt();
courses.add(courseId);
if (!courseMap.containsKey(courseId)) {
courseMap.put(courseId, new ArrayList<>());
}
courseMap.get(courseId).add(name);
}
studentMap.put(name, courses);
}
csharp
for (int i = 1; i <= k; i++) {
List<String> students = courseMap.get(i);
if (students == null) {
System.out.println(i + " 0");
continue;
}
System.out.println(i + " " + students.size());
Collections.sort(students);
for (String student : students) {
System.out.println(student);
}
}
}
}
如下代码:运行截图,效果一致:
代码如下:
import java.util.*;
public class Demo1 {
public static void main(String[] args) {
System.out.println("输入样例:");
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int k = scan.nextInt();
Map<Integer, List<String>> courseStu = new HashMap<>();
for (int i = 1; i <= k; i++) {
courseStu.put(i, new ArrayList<>());
}
for (int i = 0; i < n; i++) {
String name = scan.next();
int c = scan.nextInt();
for (int j = 0; j < c; j++) {
int course = scan.nextInt();
courseStu.get(course).add(name);
}
}
System.out.println("输出样例:");
for (int i = 1; i <= k; i++) {
List<String> stus = courseStu.get(i);
System.out.println(i + " " + stus.size());
Collections.sort(stus);
for (String stu : stus) {
System.out.println(stu);
}
}
}
}
一种比较优雅的实现方式,望采纳~~
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
Map<Integer, List<String>> map = new TreeMap<>();
for (int i = 0; i < n; i++) {
String name = sc.next();
int c = sc.nextInt();
for (int j = 0; j < c; j++) {
int course = sc.nextInt();
if (!map.containsKey(course)) {
map.put(course, new ArrayList<>());
}
map.get(course).add(name);
}
}
for (Map.Entry<Integer, List<String>> entry : map.entrySet()) {
int course = entry.getKey();
List<String> names = entry.getValue();
Collections.sort(names);
System.out.println(course + " " + names.size());
for (String name : names) {
System.out.println(name);
}
}
}
}
运行结果如下:
10 5
ZOE1 2 4 5
ANN0 3 5 2 1
BOB5 5 3 4 2 1 5
JOE4 1 2
JAY9 4 1 2 5 4
FRA8 3 4 2 5
DON2 2 4 5
AMY7 1 5
KAT3 3 5 4 2
LOR6 4 2 4 1 5
1 4
ANN0
BOB5
JAY9
LOR6
2 7
ANN0
BOB5
FRA8
JAY9
JOE4
KAT3
LOR6
3 1
BOB5
4 7
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
5 9
AMY7
ANN0
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
Process finished with exit code 0
不知道你这个问题是否已经解决, 如果还没有解决的话: