1047 Student List for Course (25 分)
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(≤), the total number of students, andK(≤), the total number of courses. ThenNlines follow, each contains a student's name (3 capital English letters plus a one-digit number), a positive numberC(≤) which is the number of courses that this student has registered, and then followed byCcourse numbers. For the sake of simplicity, the courses are numbered from 1 toK.
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:
思路:為每一個課程建立一個vector陣列,儲存選擇該門課的學生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<bits/stdc++.h> using namespace std; const int maxn=1010; vector<vector<string> > ve; bool cmp(string a,string b){ return a<b; } int main(){ int m,n; scanf("%d %d",&m,&n); ve.resize(m); for(int i=0;i<m;i++){ string st; int rig,stu; cin>>st; scanf("%d",&rig); for(int j=0;j<rig;j++){ scanf("%d",&stu); ve[stu-1].push_back(st); } } for(int i=0;i<n;i++){ printf("%d %d\n",i+1,ve[i].size()); sort(ve[i].begin(),ve[i].end(),cmp); for(int j=0;j<ve[i].size();j++){ printf("%s\n",ve[i][j].c_str()); } } return 0; }