1039 Course List for Student (25 分)_33行代码AC

立志用最少的代码做最高效的表达

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PAT甲级最优题解——>传送门

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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 (≤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.

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 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

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题意：N个查询对象， K个学院(从1开始命名)，接下来有K组数据：学院编号、学院内的学生、每个学生的名字。

最后输入N个查询对象的名字，查询他们在哪些学院， 输出名字、所在学院个数、每个学院编号(升序输出)

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分析：建立map映射，由于用名字查询，因此名字为key， 所在学院为value。 又因为学院数可能有多个，因此value为数组形式(vector)。

改进：string类型的数组做key查询起来十分耗时，可以将其散列化为int型数据存储。

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代码

#include<bits/stdc++.h> using namespace std; vector<vector<int>>hashTable((int)1e6); int strhash(const string&s) {int k = 0;for(int i = 0; i < 3; i++) {k = k*26+s[i]-'A';}k = k*26+s[3]-'0';return k; } int main() {ios::sync_with_stdio(false);int N, K; cin >> N >> K;string name; for(int i = 1; i <= K; i++) {int cour, num; cin >> cour >> num;while(num--) {cin >> name;hashTable[strhash(name)].push_back(cour);}}while(N--) {cin >> name; //不要在循环里定义string，很耗时。 int k = strhash(name);cout << name << ' ' << hashTable[k].size();sort(hashTable[k].begin(), hashTable[k].end());for(int i = 0; i < hashTable[k].size(); i++) cout << ' ' << hashTable[k][i];cout << '\n';}return 0; }

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耗时：

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