每个人都有崩溃的时候，就看你的抗压能力到底有多强，如果你的抗压能力强，有办法可以支撑到你能面对并且解决这些困难的话，你就没有问题。 ----喻言
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.
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.
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
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#include <numeric>
#include<unordered_set>
#include <climits>//INT_maxnn
//#include<bits/stdc++.h>
#define PP pair<ll,int>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
#define LL unsigned int
#define wc 1e-8
typedef long long ll;
using namespace std;
const int maxn = 26*26*26*10+10;
int n,z;
string name;
map <int,string> m;
vector<int> V[maxn];
int zh(string s)
{
int ans = 0;
for(int i = 0 ; i < 3; i++)
ans = ans *26 + s[i]-'A';
ans = ans*10 + s[3] - '0';
return ans;
}
int main()
{
scanf("%d%d",&n,&z);
for(int i = 0 ; i < n ; i++)
{
int k;
cin>>name;
scanf("%d",&k);
for(int i = 0 ; i < k ; i++)
{
int ls;
scanf("%d",&ls);
V[ls].push_back(zh(name));
m[zh(name)]=name;
}
}
vector<string> ls;
for(int i = 1 ; i <= z; i++)
{
ls.clear();
printf("%d %d\n",i,V[i].size());
for(int j = 0 ; j < V[i].size(); j++)
ls.push_back(m[V[i][j]]);
sort(ls.begin(),ls.end());
for(int j = 0 ; j < ls.size(); j++)
printf("%s\n",ls[j].c_str());
}
return 0;
}
