A1039.Course List for Student

Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists of all the courses, you are supposed to output the registered course list for each student who comes for a query.
Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (≤40,000), the number of students who look for their course lists, and K (≤2,500), the total number of courses. Then the student name lists are given for the courses (numbered from 1 to K) in the following format: for each course i, first the course index i and the number of registered students N?i?? (≤200) are given in a line. Then in the next line, N?i?? student names are given. A student name consists of 3 capital English letters plus a one-digit number. Finally the last line contains the N names of students who come for a query. All the names and numbers in a line are separated by a space.
Output Specification:

For each test case, print your results in N lines. Each line corresponds to one student, in the following format: first print the student‘s name, then the total number of registered courses of that student, and finally the indices of the courses in increasing order. The query results must be printed in the same order as input. All the data in a line must be separated by a space, with no extra space at the end of the line.
Sample Input:

11 5
4 7
BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
1 4
ANN0 BOB5 JAY9 LOR6
2 7
ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR6
3 1
BOB5
5 9
AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9

Sample Output:

ZOE1 2 4 5
ANN0 3 1 2 5
BOB5 5 1 2 3 4 5
JOE4 1 2
JAY9 4 1 2 4 5
FRA8 3 2 4 5
DON2 2 4 5
AMY7 1 5
KAT3 3 2 4 5
LOR6 4 1 2 4 5
NON9 0

題意：

　　有N個學生，K門課。現在給出選擇每門課的學生姓名，並在之後給出N個學生姓名，要求按順序給出每個學生的選課情況

參考代碼：

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 40010; //總人數
const int M = 26*26*26*10 + 1; //由姓名散列成的數字上屆
vector<int> selectCourse[M]; //每個學生選擇的課程編號

int getID(char name[]) { //hash函數，將字符串name轉換成數字
int id = 0;
for(int i = 0; i<3; i++){
id = id*26 + (name[i] - ‘A‘);
}
id = id * 10 + (name[3] - ‘0‘);
return id;
}

int main(){
char name[5];
int n,k;
scanf("%d%d",&n,&k); //人數及課程數
for(int i=0; i<k; i++){ //對每門課程
int course, x;
scanf("%d%d", &course, &x); //輸入課程編號及選課人數
for(int j = 0; j < x; j++){
scanf("%s", name); //輸入選課學生姓名
int id = getID(name); //將姓名散列為一個整數作為編號
selectCourse[id].push_back(course); //將該課程編號加入學生選擇中
}
}
for(int i = 0; i < n; i++){ //n各查詢
scanf("%s", name); //學生姓名
int id = getID(name); //獲得學生編號
sort(selectCourse[id].begin(),selectCourse[id].end()); //從小到大排序
printf("%s %d",name,selectCourse[id].size()); //姓名、選課數
for(int j = 0; j < selectCourse[id].size(); j++){
printf(" %d",selectCourse[id][j]); //選課編號
}
printf("\n");
}
return 0;
}

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

題意：
　　給出選課人數和課程數目，然後再給出每個人的選課情況，請針對每門課程輸出選課人數以及所有選該課的學生姓名

參考代碼：
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 40010; //最大的學生人數
const int maxc = 2510;  //最大課程門數

char name[maxn][5];     //maxn個學生
vector<int> course[maxc];   //course[i]存放第i門課的所有學生編號

bool cmp(int a, int b) {
    return strcmp(name[a], name[b]) < 0;    //按姓名字典序從小到大排序
}

int main() {
    int n,k,c,courseID;
    scanf("%d%d", &n, &k);  //學生人數及課程數
    for(int i = 0; i < n; i++) {
        scanf("%s %d", name[i], &c);    //學生姓名及選課數
        for(int j = 0; j < c; j++) {
            scanf("%d", &courseID);     //選擇的課程編號
            course[courseID].push_back(i);      //將學生i加入第courseID門課中
        }
    }
    for(int i = 1; i <= k; i++) {
        printf("%d %d\n", i, course[i].size());     //第i門課的學生數
        sort(course[i].begin(), course[i].end(),cmp);       //對第i門課的學生排序
        for(int j = 0; j < course[i].size(); j++) {
            printf("%s\n", name[course[i][j]]);      //輸出學生姓名
        }
    }
    return 0;
}

PAT A1039、A1047——vector常見用法