1047. Student List for Course (25)
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
分析： - 题目：给定格式化的人名ABC0,和课号(1-N),给出选课情况，升序输出每个课程的选课人名(升序)。 - 利用stl的sort排序完成，但string的排序太慢，导致最后的测试点超时。故改变思路，string to int 中间用数字表示人名，排完序后，再转换回来,这样人名排序就是数字的比较。26进制10进制转化。
code:
#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
int nameToNum(
string name)
{
int num=(name[
0]-
'A');
num*=
26;
num+=(name[
1]-
'A');
num*=
26;
num+=(name[
2]-
'A');
num*=
26;
num+=(name[
3]-
'0');
return num;
}
string numToName(
int n)
{
char tmp[
5];
tmp[
4]=
'\0';
tmp[
3]=
char(n%
26+
'0');
n/=
26;
tmp[
2]=
char(n%
26+
'A');
n/=
26;
tmp[
1]=
char(n%
26+
'A');
n/=
26;
tmp[
0]=
char(n%
26+
'A');
return tmp;
}
vector<int> course[
2550];
int main()
{
freopen(
"in",
"r",stdin);
int N,K,C,tmp,iname;
char name[
5];
scanf(
"%d%d",&N,&K);
for(
int i=
0;i<N;i++)
{
scanf(
"%s%d",name,&C);
iname=nameToNum(name);
for(
int j=
0;j<C;j++)
{
scanf(
"%d",&tmp);
course[tmp].push_back(iname);
}
}
int size=
0;
string sname;
for(
int i=
1;i<=K;i++)
{
size=course[i].size();
printf(
"%d %d\n",i,size);
sort(course[i].begin(),course[i].end());
for(
int j=
0;j<course[i].size();j++)
{
sname=numToName(course[i][j]);
printf(
"%s\n",sname.c_str());
}
}
return 0;
}
AC
