Course List for Student

题目来源

http://codeup.cn/problem.php?cid=100000596&pid=0

题目描述

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.

输入

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 Ni (<= 200) are given in a line. Then in the next line, Ni 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.

输出

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.

样例输入

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

样例输出 

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

题目的主要难点是对每个学生姓名的处理，这里用到的是一个散列处理，也就是把每个学生的姓名映射到一个整数，为了保证这个整数是唯一的，常见的处理方法是进制的转换。
如果把大学字母A表示为0，则Z对应的就是25，这样以来每个字母对应的数字永远不会超过25，所以可以把学生的姓名看成是一个26进制的数，接下来把这个26进制数转换成10进制的方便我们处理。
PS：一定要记得把规模大的数组设置为全局，不然会运行错误。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<set>
#include<string>
#include<string.h>
#include<vector>
using namespace std;
int change(string s) {
	int ans = 0;
	for(int i = 0; i < s.size() - 1; i++) {
		ans = ans * 26 + (s[i] - 'A');
	}
	return ans * 10 + s[3] - '0';
}
vector<int> course[25 * 26 * 26 * 10 + 25 * 26 * 10 + 25 * 10 + 10]; 
int main() {
	int N, K;
	string s;
	cin >> N >> K;
	int a, b;
	for(int i = 0; i < K; i++) {
		cin >> a >> b;
		for(int j = 0; j < b; j++) {
			cin >> s;
			course[change(s)].push_back(a);
		}
	}
	for(int i = 0; i < N; i++) {
		cin >> s;
		cout << s << " ";
		cout << course[change(s)].size();
		sort(course[change(s)].begin(), course[change(s)].end());
		for(vector<int>::iterator it = course[change(s)].begin(); it != course[change(s)].end(); it++) {
			cout << " " << *it;
		}
		cout << endl;
	}
	return 0;
}