【PAT甲级】1028 List Sorting

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📝原题地址：题目详情 - 1028 List Sorting (pintia.cn)
🔑中文翻译：列表排序
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1028 List Sorting

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input Specification:

Each input file contains one test case. For each case, the first line contains two integers N (≤10⁵) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student’s record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).

Output Specification:

For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID’s; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID’s in increasing order.

Sample Input 1:

3 1
000007 James 85
000010 Amy 90
000001 Zoe 60
1
2
3
4

Sample Output 1:

000001 Zoe 60
000007 James 85
000010 Amy 90
1
2
3

Sample Input 2:

4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98
1
2
3
4
5

Sample Output 2:

000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60
1
2
3
4

Sample Input 3:

4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 9
1
2
3
4
5

Sample Output 3:

000002 James 9
000001 Zoe 60
000007 James 85
000010 Amy 90
1
2
3
4

题意

第一行给定学生人数 N 和序号 C 。

按照给定序号数对学生进行排序：

如果 C = 1 ，则按照 ID 升序的顺序排序。

如果 C = 2 ，则按照名称以不降序的顺序排序。

如果 C = 3 ，则按照成绩以不降序的顺序排序。

当出现学生名字相同或是成绩相同的情况时，按照 ID 升序的顺序，对他们进行排序。

思路

1. 用结构体数组存储学生信息，注意要用 scanf 输入，这题数据量比较大，用 cin 会超时。
2. 分别设定三个比较函数，然后根据给定的 C 值进行对应的排序。
3. 输出排序后的结果。

代码

#include
using namespace std;

const int N = 100010;
struct Student {
    string id, name;
    int grade;
}all[N];

//按照id升序排序
bool cmp1(Student& s1, Student& s2)
{
    return s1.id < s2.id;
}

//按照名字升序排序
bool cmp2(Student& s1, Student& s2)
{
    if (s1.name != s2.name)    return s1.name < s2.name;
    return s1.id < s2.id;
}

//按照成绩升序排序
bool cmp3(Student& s1, Student& s2)
{
    if (s1.grade != s2.grade)  return s1.grade < s2.grade;
    return s1.id < s2.id;
}

int main()
{
    int n, c;
    scanf("%d %d", &n, &c);

    //输入学生信息
    char id[10], name[10];
    for (int i = 0; i < n; i++)
    {
        int grade;
        scanf("%s %s %d", &id, &name, &grade);
        all[i] = { id,name,grade };
    }

    if (c == 1)    sort(all, all + n, cmp1);   //按照id升序排序
    else if (c == 2)   sort(all, all + n, cmp2);   //按照名字升序排序
    else    sort(all, all + n, cmp3);   //按照成绩升序排序

    //输出排序结果
    for (int i = 0; i < n; i++)
        printf("%s %s %d\n", all[i].id.c_str(), all[i].name.c_str(), all[i].grade);

    return 0;
}
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