PAT A 1039. Course List for Student (25)

题目

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 (<=40000), the number of students who look for their course lists, and K (<=2500), 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 54 7BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE11 4ANN0 BOB5 JAY9 LOR62 7ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR63 1BOB55 9AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9

Sample Output:

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

 

比较容易想到用map建立名字和id的对应关系，然后暂存每个人的课程，再查找输出。

实测会超时，因为以红黑树实现的map每次名字输入的查找代价过大log(n)。

解决：由于名字取值的特殊性（3个字母加1个数字），可由名字抓换得唯一的id，id的范围不算太大，这样查找的代价可降至常数级别。

 

代码：

#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;int Get_id(char name[5]);//将名字转换为id，（由于名字取值的限制，可以转换）int main(){int n,k;cin>>n>>k;int id=0;vector<int> *course;//课程信息course=new vector<int>[200000];//由于名字限制，范围小于26*26*26*10,直接根据id取相应的编号，可以大幅节约时间int i,j,num_name,cur_id;char name_tmp[5];for(i=0;i<k;i++){scanf("%d",&cur_id);scanf("%d",&num_name);for(j=0;j<num_name;j++){scanf("%s",name_tmp);id=Get_id(name_tmp);course[id].push_back(cur_id);}}for(i=0;i<200000;i++)//各个id排序sort(course[i].begin(),course[i].end());int num_cur;for(i=0;i<n;i++)//输入名字，转换id，输出相应的{scanf("%s",name_tmp);printf("%s ",name_tmp);id=Get_id(name_tmp);num_cur=course[id].size();printf("%d",num_cur);for(j=0;j<num_cur;j++)printf(" %d",course[id][j]);printf("\n");}return 0;}int Get_id(char name[5]){int id;id=(name[0]-'A')*10*26*26+(name[1]-'A')*10*26+(name[2]-'A')*10+name[3]-'0';return id;}