PAT 1028 List Sorting (25分) 用char[]，不要用string

题目

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 (≤10^​5 ) 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

Sample Output 1:

000001 Zoe 60

000007 James 85

000010 Amy 90

Sample Input 2:

4 2

000007 James 85

000010 Amy 90

000001 Zoe 60

000002 James 98

Sample Output 2:

000010 Amy 90

000002 James 98

000007 James 85

000001 Zoe 60

Sample Input 3:

4 3

000007 James 85

000010 Amy 90

000001 Zoe 60

000002 James 90

Sample Output 3:

000001 Zoe 60

000007 James 85

000002 James 90

000010 Amy 90

题目解读

给出N个学生信息和一个数字C，每个学生的信息包括ID（6位数字），姓名（长度最多为8的字符串，无空格），分数（0-100）。根据数字C的取值，对学生信息按照不同策略进行排序，最终输出排序后的学生信息：

C=1：按 ID 递增；

C=2：按姓名递增，如果同名，按ID递增；

C=3：按分数递增，如果同分，按ID递增。

思路分析

这不就是三种排序策略就完了吗？

• 首先创建结构体 Student 保存学生信息，注意 name 字段不要用string！！！，否则你最后一个测试点会是 运行超时，实际是内存溢出！
  
  
  
  其实可以想想，题目给出说明姓名字段长度不超过10个字符，怎么可能没用呢，是吧！
• 如何排序，因为我们使用sort()函数对整个结构体数组进行排序，自己实现的比较函数只能是这样int cmp(Student a, Student b)，所以我们要把 C 定义成一个全局变量，然后在这个函数中根据 C的取值进行不同的逻辑实现，当然你也可以实现三个比较函数，但我觉得没有必要。

代码

题目比较简单，代码注释也挺详细，没什么好说的。有个地方需要注意：学号是6位数字，输出必须用printf("%06d", id)格式化。

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

// 学生信息
struct Student {
    int id, score;
    // string name;
    char name[10];
}stu[10001];

// 根据那个字段排序
int flag;

// 自定义比较函数
int cmp(Student a, Student b) {
    // 按照ID递增
    if (flag == 1) return a.id < b.id;
    // 按照姓名自增
    else if (flag == 2) {
        // 重名就比较ID
        if (strcmp(a.name, b.name) == 0) return a.id < b.id;
        return strcmp(a.name, b.name) <= 0;
    } else {
        // 按照分数递增，相同就比较ID
        return a.score != b.score ? a.score < b.score : a.id < b.id;
    }
}

int main() {

    // N 个学生
    int n;
    // 根据哪个字段排序
    cin >> n >> flag;
    // 读入学生信息
    for (int i = 0; i < n; ++i) {
        // cin >> stu[i].id >> stu[i].name >> stu[i].score;
        scanf("%d %s %d", &stu[i].id, stu[i].name, &stu[i].score);
    }
    // 排序
    sort(stu, stu + n, cmp);
    // 输出
    for (int i = 0; i < n; ++i) {
        // printf("%06d ", stu[i].id);
        // cout << stu[i].name << " " << stu[i].score << endl;
        printf("%06d %s %d\n", stu[i].id, stu[i].name, stu[i].score);
    }

    return 0;
}