PAT_A 1047. Student List for Course (25)

1047. Student List for Course (25)

Zhejiang University has 40000 students and provides 2500 courses. Now given the registeredcourse 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 Nlines follow, each contains a student's name (3 capital English letters plus a one-digitnumber), 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 5ZOE1 2 4 5ANN0 3 5 2 1BOB5 5 3 4 2 1 5JOE4 1 2JAY9 4 1 2 5 4FRA8 3 4 2 5DON2 2 4 5AMY7 1 5KAT3 3 5 4 2LOR6 4 2 4 1 5Sample Output:1 4ANN0BOB5JAY9LOR62 7ANN0BOB5FRA8JAY9JOE4KAT3LOR63 1BOB54 7BOB5DON2FRA8JAY9KAT3LOR6ZOE15 9AMY7ANN0BOB5DON2FRA8JAY9KAT3LOR6ZOE1

• 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');//懒得写循环了，之前是写的循环    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