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  • PAT1047: Student List for Course

    1047. Student List for Course (25)

    时间限制
    400 ms
    内存限制
    64000 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    Zhejiang University has 40000 students and provides 2500 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 (<=40000), the total number of students, and K (<=2500), 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

    思路
    类似PAT1039.用string仍然会超时,看来对string的各种操作要比直接操作char字符数组更慢。
    超时代码
    #include<iostream>
    #include<vector>
    #include<algorithm>
    #include<string>
    using namespace std;
    bool cmp(const string& a,const string& b)
    {
        return a.compare(b) < 0;
    }
    int main()
    {
       int N,K;
       while(cin >> N >> K)
       {
           vector<vector<string>> courses(K + 1);
           courses.clear();
           for(int i = 0; i < N;i++)
           {
               string name;
               int n;
               cin >> name >> n;
               for(int j = 0;j < n;j++)
               {
                   int course;
                   cin >> course;
                   courses[course].push_back(name);
               }
           }
    
    
           for(int i = 1;i <= K;i++)
           {
               cout << i << " " << courses[i].size() << endl;
               sort(courses[i].begin(),courses[i].end(),cmp);
               for(int j = 0;j < courses[i].size();j++)
                  cout << courses[i][j] << endl;
           }
       }
    }
    AC代码
    #include <cstdio>
    #include <vector>
    #include <algorithm>
    #include <string.h>
    using namespace std;
    char name[40010][5];
    vector<int> course[2510];
    bool cmp1(int a, int b) {
        return strcmp(name[a], name[b]) < 0;
    }
    int main() {
        int n, k;
        scanf("%d %d", &n, &k);
        for(int i = 0; i < n; i++) {
            int cnt, temp;
            scanf("%s %d", name[i], &cnt);
            for(int j = 0; j < cnt; j++) {
                scanf("%d", &temp);
                course[temp].push_back(i);
            }
        }
        for(int i = 1; i <= k; i++) {
            printf("%d %d
    ", i, course[i].size());
            sort(course[i].begin(), course[i].end(), cmp1);
            for(int j = 0; j < course[i].size(); j++) {
                printf("%s
    ", name[course[i][j]]);
            }
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/0kk470/p/7881363.html
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