PAT A1028. List Sorting (25)

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本文介绍了一个简单的程序设计问题，即如何模仿Excel的排序功能。该程序读取学生的记录（包括ID、姓名和成绩），并根据指定的列对学生记录进行排序。提供了完整的C++代码实现，演示了如何使用比较函数来实现不同条件下的排序。

题目链接

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

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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

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

此题属于简单题，只是需要对cmp函数进行活用即可，代码如下：

#include<cstdio>
#include <algorithm>
#include<cstring>
using namespace std;
struct Stu {
    int id;
    char name[10];
    int score;
};
Stu stu[100005];
int c;
bool cmp(Stu a, Stu b) {
    int s;
    switch (c) {
    case 1:return a.id < b.id;
    case 2: s = strcmp(a.name, b.name);
        if (s)return s < 0;
        else return a.id < b.id; 
    case 3: if (a.score != b.score)return a.score < b.score;
            else return a.id < b.id; 
    }
}
int main() {
    int n;
    scanf("%d %d", &n, &c);
    for (int i = 0; i < n; i++) {
        scanf("%d %s %d", &stu[i].id, stu[i].name, &stu[i].score);
    }
    sort(stu, stu + n, cmp);
    for (int i = 0; i < n; i++) {
        printf("%06d %s %d\n", stu[i].id, stu[i].name, stu[i].score);
    }
    return 0;
}