立志用最少的代碼做最高效的表達

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PAT甲級最優題解——>傳送門

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Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course 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 (≤40,000), the total number of students, and K (≤2,500), the total number of courses. Then N lines follow, each contains a student’s name (3 capital English letters plus a one-digit number), 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 5
ZOE1 2 4 5
ANN0 3 5 2 1
BOB5 5 3 4 2 1 5
JOE4 1 2
JAY9 4 1 2 5 4
FRA8 3 4 2 5
DON2 2 4 5
AMY7 1 5
KAT3 3 5 4 2
LOR6 4 2 4 1 5

Sample Output:
1 4
ANN0
BOB5
JAY9
LOR6
2 7
ANN0
BOB5
FRA8
JAY9
JOE4
KAT3
LOR6
3 1
BOB5
4 7
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
5 9
AMY7
ANN0
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1

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題意：第一行輸入學生人數、學院個數。 接下來輸入學生名， 輸入其加入的學院個數， 以及學院編號。 最后按學院編號輸出每個學院的學生(學生升序排序)

算法設計：采用STL嵌套即可。 但如果直接以名字為參數存放在數組里會導致超時(存儲字符串很耗時)，因此可以考慮將字符串散列化、為每個字符串設計一個獨特的int型編碼，存儲編碼即可。

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儲備知識擴展(很重要，提高效率，降低碼量)：

• set——有序去重集合。

• map——有序去重映射

• multiset——有序不去重集合

• multimap——有序不去重映射

• unordered_set——無序不去重集合(普通集合)

• unordered_map——無序不去重映射(普通映射)

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#include<bits/stdc++.h> using namespace std;int incode(string s) {int x = 0, i;for(i = 0; i < s.size()-1; i++) x = x*26 + s[i]-'A';x = x*26 + s[i]-'0';return x; }int main() {ios::sync_with_stdio(false);int n, num_cour; cin >> n >> num_cour;map<int, vector<int> >m; //key是學院號，value是名字的散列化 unordered_map<int, string>um; //字符串與其散列化的映射 for(int i = 0; i < n; i++) {string s; int num; cin >> s >> num;int in = incode(s); um[in] = s;for(int i = 0; i < num; i++) {int x; cin >> x;m[x].push_back(in);}} for(int i = 1; i <= num_cour; i++) {cout << i << ' ' << m[i].size() << '\n';sort(m[i].begin(), m[i].end());for(auto j : m[i]) cout << um[j] << '\n';}return 0; }

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耗時：

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求贊哦~ (?ω?)

總結

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