PAT-1047. Student List for Course (25)

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

---

 

使用string会超时，使用char时，char[]和char*的转换也很复杂，还没有很好的办法。由于所给字符串是有规律的，我考虑将字符串hash成数字，存储比较起来省时；

hash就是将每一位的字符转换成对应的数值（字符值*位数），高位字符对应数值大，方便排序。但hash不是最关键的部分，最重要的部分是char*和char[]的转换，我现在还没有很好的解决方法。

当然由于vector<char[]>是不符合语法的，我们可以考虑构建结构体，将char[]放入结构体，这样也避免了使用char*的问题。

以上两种方法均有实现。（注意当char数组存放4位字符时，要定义char[5]，最后一位需要''）

根据运行时间知，使用hash确实效率更高一些，将两种方法结合使用是最完美的。

#include <bits/stdc++.h>

using namespace std;

int N, C;
vector<int> course[40004];

int hashName(char* name) {
    int hName = 0;
    hName += (name[0] - 'A') * 26 * 26 * 10;
    hName += (name[1] - 'A') * 26 * 10;
    hName += (name[2] - 'A') * 10;
    hName += (name[3] - '0');
    return hName;
}

void printName(int hName) {
    char name[5];
    name[0] = hName/10/26/26 + 'A';
    name[1] = (hName%(10*26*26))/10/26 + 'A';
    name[2] = (hName%(10*26))/10 + 'A';
    name[3] = hName%(10) + '0';
    printf("%s
", name);
}

int main()
{
    cin>> N>> C;
    char name[5];
    //name.resize(4);
    int m, c, k;
    for(int i = 0; i < N; i++) {
        //cin>> name;
        scanf("%s", name);
        //printf("name=%s
", name);
        int hName = hashName(name);
        scanf("%d", &k);
        for(int j = 0; j < k; j++) {
            scanf("%d", &c);
            course[c].push_back(hName);
        }
    }
    for(int i = 1; i <= C; i++) {
        printf("%d %d
", i, course[i].size());
        sort(course[i].begin(), course[i].end());
        //char* name = name
        for(int j = 0; j < course[i].size(); j++) {
            printName(course[i][j]);
            //printf("%s
", hname(course[i][j]));
        }
    }
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

struct Student {
    char name[5];
    bool operator < (const Student s) const {
        return strcmp(name, s.name) < 0;
    }
};

int N, C;
vector<Student> course[40004];

int main()
{
    cin>> N>> C;
    Student stu;
    //name.resize(4);
    int m, c, k;
    for(int i = 0; i < N; i++) {
        //cin>> name;
        scanf("%s", stu.name);
        scanf("%d", &k);
        for(int j = 0; j < k; j++) {
            scanf("%d", &c);
            course[c].push_back(stu);
        }
    }
    for(int i = 1; i <= C; i++) {
        printf("%d %d
", i, course[i].size());
        sort(course[i].begin(), course[i].end());
        //char* name = name
        for(int j = 0; j < course[i].size(); j++) {
            printf("%s
", course[i][j].name);
            //printf("%s
", hname(course[i][j]));
        }
    }
    return 0;
}