PAT——StudentListforCourse（注意运行超时问题）

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
结尾无空行


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
结尾无空行

AC代码

#include
using namespace std;
int main()
{map<int,vector<string> > maps;int n,k;cin>>n>>k; for(int i&＃61;0;i<n;i&＃43;&＃43;){string s;cin>>s;int num;cin>>num;while(num--){int t;
// cin>>t;scanf("%d",&t);maps[t].push_back(s);}}for(int i&＃61;1;i<&＃61;k;i&＃43;&＃43;){cout<<i<<" "<<maps[i].size()<<endl;sort(maps[i].begin(),maps[i].end());for(int j&＃61;0;j<maps[i].size();j&＃43;&＃43;)
// cout<printf("%s\n",maps[i][j].c_str());}
}

注意

如果最后一个测试点运行超时&＃xff0c;可以试试将cin&＃xff0c;cout改为scanf和printf

在此之上又有一个问题就是如何输出string类型的数据&＃xff0c;需要进行如下处理&＃xff0c;&＃xff08;具体可见这篇文章——printf输出string类型&＃xff09;

printf("%s\n",maps[i][j].c_str());