1047 Student List for Course

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 (≤), the total number of students, and K (≤), 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 (≤) 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

题意：

　　给出一组学生的选课清单，让你给出选这门课的学生的名单。

思路：

　　这是一道简单的排序题，但是最后一组数据会卡超时。所以要使用scanf和printf来输入，输出。使用scanf输入的时候，要使用char[]。

Code：

#include <bits/stdc++.h>

using namespace std;

char name[40005][5];
bool cmp(int a, int b) {
    if (strcmp(name[a], name[b]) < 0)
        return true;
    else
        return false;
}

int main() {
    int n, k, c, t;
    scanf("%d%d", &n, &k);
    vector<vector<int> > v(k + 1);
    for (int i = 0; i < n; ++i) {
        scanf("%s%d", name[i], &c);
        for (int j = 0; j < c; ++j) {
            scanf("%d", &t);
            v[t].push_back(i);
        }
    }
    for (int i = 1; i <= k; ++i) {
        printf("%d %d
", i, v[i].size());
        sort(v[i].begin(), v[i].end(), cmp);
        for (int j = 0; j < v[i].size(); ++j) printf("%s
", name[v[i][j]]);
    }
    return 0;
}