Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤), the total number of students, and K (≤), the total number of courses. Then N lines follow, each contains a student's name (3 capital English letters plus a one-digit number), a positive number C (≤) which is the number of courses that this student has registered, and then followed by C course numbers. For the sake of simplicity, the courses are numbered from 1 to K.
Output Specification:
For each test case, print the student name lists of all the courses in increasing order of the course numbers. For each course, first print in one line the course number and the number of registered students, separated by a space. Then output the students' names in alphabetical order. Each name occupies a line.
Sample Input:
10 5ZOE1 2 4 5ANN0 3 5 2 1BOB5 5 3 4 2 1 5JOE4 1 2JAY9 4 1 2 5 4FRA8 3 4 2 5DON2 2 4 5AMY7 1 5KAT3 3 5 4 2LOR6 4 2 4 1 5
Sample Output:
1 4ANN0BOB5JAY9LOR62 7ANN0BOB5FRA8JAY9JOE4KAT3LOR63 1BOB54 7BOB5DON2FRA8JAY9KAT3LOR6ZOE15 9AMY7ANN0BOB5DON2FRA8JAY9KAT3LOR6ZOE1
1 #include2 #include 3 #include 4 #include 5 //题目不难，不过数据量很大，很容易超时。可以定义一个string数组name按读入顺序存储读取的名字字符串，然后定义一个长度为K + 1的vector 的数组course，其数组下标存储课程号，数组元素存储选修该课程的学生的名字在name中的下标。在输出时对course中的数组元素按要求排序，然后输出即可。 6 //注意点 7 //排序时直接对字符串进行排序非常消耗时间，可以利用字符串的下标代替字符串本身进行排序 8 9 using namespace std;10 int N, K, C;11 int main()12 {13 cin >> N >> K;14 vector >courses(K + 1);15 vector name(N + 1);16 int num;17 for (int i = 0; i < N; ++i)18 {19 cin >> name[i] >> C;20 for (int j = 0; j < C; ++j)21 {22 cin >> num;23 courses[num].push_back(i);24 }25 }26 for (int i = 1; i <= K; ++i)27 {28 if (courses[i].size() == 0)continue;29 cout << i << " " << courses[i].size() << endl;30 sort(courses[i].begin(), courses[i].end(), [=](int a, int b) { return name[a] < name[b]; });31 for (auto a : courses[i])32 cout << name[i] << endl;33 }34 return 0;35 }