Zhejiang University has 40000 students and provides 2500 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 (<=40000), the total number of students, and K (<=2500), 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
题目大意：给出每个学生选的课程， 输出选择课程的学生， 升序输出
注意点： 保存学生的名字最小应该为char ch[5]
思路： 因为学生的名字是固定格式的， 为了减少时间， 把学生的名字hash成整数， 用char stud[18000][5]记录学生的名字, 第一个下标即学生名字对应的hash值
把每个学生的hash值添加到课程上， 对每一门课程的学生进行排序后输出
1 #include<stdio.h> 2 #include<algorithm> 3 #include<vector> 4 #include<string.h> 5 using namespace std; 6 vector<int> course[2501]; 7 bool cmp(int a,int b){return a<b;} 8 char stud[180000][5]; 9 int main() 10 { 11 int i,j,n,m,k,t,id; 12 char name[5]; 13 scanf("%d %d",&n,&m); 14 for(i=0;i<n;i++) 15 { 16 scanf("%s %d",name,&k); 17 id = (name[0]-'A')*26*26*10+(name[1]-'A')*26*10+(name[2]-'A')*10+name[3]-'0'; 18 strcpy(stud[id],name); 19 for(j=0;j<k;j++) 20 { 21 scanf("%d",&t); 22 course[t].push_back(id); 23 } 24 } 25 for(i=1;i<=m;i++) 26 { 27 printf("%d %d ",i,course[i].size()); 28 sort(course[i].begin(),course[i].end(),cmp); 29 for(j=0;j<course[i].size();j++) 30 printf("%s ",stud[course[i][j]]); 31 } 32 return 0; 33 }