题目

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.

输入

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.

输出

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.

样例输入 

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

样例输出 

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

题意理解

给你n个人，k门课

每个人选了几门课和具体的课是什么

输出k门课从小到大

然后每门课里面有几个人选了

输出这几个人的名字 注意按字典序升序输出

这题第四个测试点用cin cout是过不去的 加入读入优化也过不去

然后哈希表里面存的值最好用vector吧 用vector然后再去每个课对应的学习排个序

如果用存的值是set的话，第四个点还是会超时。

总结就是哈希查表 然后读入读出都用scanf 和 printf 吧

代码 

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef unordered_map<int,vector<string>> MSS;
int n,k;
MSS ma;
int main(){
//     ios::sync_with_stdio(0);
//     cin.tie(0);cout.tie(0);
    scanf("%d%d",&n,&k);
    string a;
    a.resize(5);
    while(n--){
        int nu;
        scanf("%s",&a[0]);
        scanf("%d",&nu);
        while(nu--){
            int x;
            scanf("%d",&x);
            ma[x].push_back(a);
        }
    }
    for(int i=1;i<=k;i++){
        sort(ma[i].begin(),ma[i].end());
    }
    for(int i=1;i<=k;i++){
        printf("%d %d\n",i,ma[i].size());
        for(auto tmp:ma[i]){
            printf("%s\n",tmp.c_str());
        }
    }
    return 0;
}