1047 Student List for Course (25)

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:

2. 10 5
3. ZOE1 2 4 5
4. ANN0 3 5 2 1
5. BOB5 5 3 4 2 1 5
6. JOE4 1 2
7. JAY9 4 1 2 5 4
8. FRA8 3 4 2 5
9. DON2 2 4 5
10. AMY7 1 5
11. KAT3 3 5 4 2
12. LOR6 4 2 4 1 5

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Sample Output:

2. 1 4
3. ANN0
4. BOB5
5. JAY9
6. LOR6
7. 2 7
8. ANN0
9. BOB5
10. FRA8
11. JAY9
12. JOE4
13. KAT3
14. LOR6
15. 3 1
16. BOB5
17. 4 7
18. BOB5
19. DON2
20. FRA8
21. JAY9
22. KAT3
23. LOR6
24. ZOE1
25. 5 9
26. AMY7
27. ANN0
28. BOB5
29. DON2
30. FRA8
31. JAY9
32. KAT3
33. LOR6
34. ZOE1

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标题大意：给出选课人数和课程数量，然后再给出每个人的选课环境，请针对每门课程输出选课人数以及全部选该课的弟子姓名，按照字典序输出。
分析：用vector存储选课的弟子，string纪录弟子的姓名。

2. #include<algorithm>
3. #include <iostream>
4. #include  <cstdlib>
5. #include  <cstring>
6. #include   <string>
7. #include   <vector>
8. #include   <cstdio>
9. #include    <queue>
10. #include    <stack>
11. #include    <ctime>
12. #include    <cmath>
13. using namespace std;
15. int main(void)
16. {
17.     #ifdef test
18.     freopen("in.txt","r",stdin);
19.     //freopen("in.txt","w",stdout);
20.     clock_t start=clock();
21.     #endif //test
23.     int n,k;scanf("%d%d",&n,&k);
24.     vector<string>s[5+k];
25.     for(int i=0;i<n;++i)
26.     {
27.         string ss;cin>>ss;
28.         int cnt;scanf("%d",&cnt);
29.         for(int j=0;j<cnt;++j)
30.         {
31.             int a;scanf("%d",&a);
32.             s[a].push_back(ss);
33.         }
34.     }
35.     for(int i=1;i<=k;++i)
36.     {
37.         printf("%d %lu\n",i,s[i].size());
38.         if(s[i].size())
39.         {
40.             sort(s[i].begin(),s[i].end());
41.             for(auto it=s[i].begin();it!=s[i].end();++it)
42.                 cout<<*it<<'\n';
43.         }
44.     }
46.     #ifdef test
47.     clockid_t end=clock();
48.     double endtime=(double)(end-start)/CLOCKS_PER_SEC;
49.     printf("\n\n\n\n\n");
50.     cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
51.     cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
52.     #endif //test
53.     return 0;
54. }

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