PAT 甲級 A1039

1039 Course List for Student (25分)

題目描述

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.

輸入格式

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (≤40,000), the number of students who look for their course lists, and K (≤2,500), 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.

輸出格式

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 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

Sample Output:

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

總結

  看一眼題目要求查詢得到這麼複雜的結果，就知道…肯定要用散列。剛開始我是用的C++11的unordered_map，後來發現輸出結果不對…看來沒掌握到精髓，於是就自己寫了個散列函數。散列函數的實現也很簡單,每移一個位，就乘一個位不同情況的個數(類似於計算二進制)。再配合一個變長數組vector，就可以計算出每個學生和他的對應選課。思路和實現都在代碼下面了。

AC代碼

#include <iostream>
#include<algorithm>
#include<vector>
using namespace std;
vector<int> stu[26 * 26 * 26 * 10 + 1];
int strHash(char stu[]) {  //散列,三個大寫字母的和一個數字的組合最多爲26*26*26*10 < 270000
	int id = 0;
	for (int i = 0; i < 3; i++) id = id * 26 + (stu[i] - 'A');
	id = id * 10 + stu[3] - '0';
	return id;
}
int main() {
	int n, k, cousreId, stuNums;
	char temp[5];
	scanf("%d %d", &n, &k);
	for (int i = 0; i < k; i++) {
		scanf("%d %d", &cousreId, &stuNums);
		for (int j = 0; j < stuNums; j++) {
			scanf("%s", temp);
			int index = strHash(temp);
			stu[index].push_back(cousreId);
		}
	}
	while (~scanf("%s", temp)) {
		int index = strHash(temp);
		printf("%s %d", temp, stu[index].size());
		if (stu[index].size()) {
			sort(stu[index].begin(), stu[index].end());
			for (auto i = stu[index].begin(); i != stu[index].end(); i++)
				printf(" %d", *i);
		}
		printf("\n");
	}
	return 0;
}