1047 Student List for Course （25 point(s)）

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

经验总结：

这一题，和​​PAT 甲级 1039 Course List for Student​​  很像，核心问题都是对于名字的处理，如果使用cin和cout，最后一个测试点就会超时，对于名字的处理当然是将其转换成一个整数，在输出时再将其逆向转换为字符串就行了，问题不大(๑•̀ㅂ•́)و✧~~

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <vector>
using namespace std;
int convert(char a[])
{
  int ans=0;
  for(int i=0;i<3;++i)
    ans=ans*26+a[i]-'A';
  ans=ans*10+a[3]-'0';
  return ans;
}
void reconvert(int x,char a[])
{
  a[4]='\0';
  a[3]=x%10+'0';
  x/=10;
  for(int i=2;i>=0;--i)
  {
    a[i]=x%26+'A';
    x/=26;
  }
}
int main()
{
  int n,m,a,b,k;
  scanf("%d %d",&n,&k);
  vector<vector<int> > ans;
  char str[7];
  ans.resize(k+1);
  for(int i=0;i<n;++i)
  {
    scanf("%s",str);
    int temp=convert(str);
    scanf("%d",&m);
    for(int j=0;j<m;++j)
    {
      scanf("%d",&a);
      ans[a].push_back(temp);
    }
  }
  for(int i=1;i<=k;++i)
  {
    printf("%d %d\n",i,ans[i].size());
    sort(ans[i].begin(),ans[i].end());
    for(int j=0;j<ans[i].size();++j)
    {
      reconvert(ans[i][j],str);
      printf("%s\n",str);
    }
  }
  return 0;
}