1047 Student List for Course (25 分)

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 (≤40,000), the total number of students, and K (≤2,500), 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 (≤20) 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 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

Sample Output:

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

题目大意：

         给出n个学生以及课程数k。随之是n行信息，每一行包括了学生名字以及他的选课数c，接着便是相应的c个课程编号。要求按照课程的编号整理出每门课包含的学生数以及相应的学生名字（字典序升序）

 

思路：

     课程数组course，存储每个学生的名字，每次录入信息时， 对应的course[i]，就存入当前学生的名字，输出时，从小到大输出，并用sort排序好的就行了。

    

参考代码：

#include<vector> #include<iostream> #include<algorithm> using namespace std; int n, k, c, idx; vector<vector<string> > course; bool cmp(string &a, string &b){return a < b;} int main(){ scanf("%d%d", &n, &k); course.resize(k + 1); for(int i = 0; i < n; ++i){ string name; cin >> name >> c; for(int j = 0; j < c; ++j){ scanf("%d", &idx); course[idx].push_back(name); } } for(int i = 1; i <= k; ++i){ printf("%d %d\n", i, course[i].size()); sort(course[i].begin(), course[i].end(), cmp); for(int j = 0; j < course[i].size(); ++j) printf("%s\n", course[i][j].c_str()); } return 0; }