1028. List Sorting (25)-优快云博客

本文解析PAT-A-1028题目，使用C语言实现学生记录的排序功能。根据输入的学生记录（包括ID、姓名和成绩），按照指定列进行排序，并给出排序后的结果。代码使用qsort函数实现不同条件下的排序。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题地址：http://www.patest.cn/contests/pat-a-practise/1028

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

解题思路：考察STL中sort用法。但定睛一看我当时竟然是用qsort，还是太菜了

代码如下：

#include <stdio.h>
#include <algorithm>
#include <cstring>
#define MAXN 100010
typedef struct record{
    char id[8];
    char name[10];
    int grade;  
}rec;

rec st[MAXN];
int N,C;

int cmp1(const void* _a, const void* _b){
    rec a = *(rec*)_a;
    rec b = *(rec*)_b;
    return strcmp(a.id,b.id);
}

int cmp2(const void* _a, const void* _b){
    rec a = *(rec*)_a;
    rec b = *(rec*)_b;
    int t = strcmp(a.name,b.name);
    if(t == 0) return strcmp(a.id,b.id);
    else return t;
}

int cmp3(const void* _a, const void* _b){
    rec a = *(rec*)_a;
    rec b = *(rec*)_b;
    int t = a.grade - b.grade;
    if(t > 0) return 1;
    else if(t < 0) return -1;
    else return strcmp(a.id,b.id);
}

int main(){
    freopen("1028.txt","r",stdin);
    scanf("%d%d",&N,&C);
    for(int i = 0; i < N; i++){
        scanf("%s%s%d",&st[i].id,&st[i].name,&st[i].grade); 
    }
    switch(C){
        case 1: qsort(st,N,sizeof(st[0]),cmp1); break;
        case 2: qsort(st,N,sizeof(st[0]),cmp2); break;
        case 3: qsort(st,N,sizeof(st[0]),cmp3); break;  
    }
    for(int i = 0; i < N; i++){
        printf("%s %s %d\n",st[i].id,st[i].name,st[i].grade);   
    }
    return 0;
}