PAT 1028. List Sorting (25) 重写sort 函数中的cmp

最新推荐文章于 2022-05-20 15:51:44 发布

原创最新推荐文章于 2022-05-20 15:51:44 发布 · 798 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一个模拟Excel排序功能的算法实现，可以根据不同列对学生记录进行排序，包括ID、姓名及成绩，并提供了具体的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

此题要注意两点：一是结构体数组要设置为全局变量，否则会overflow；另一点就是sort()中cmp函数的重写方法。

具体实现代码如下：

#include<cstdio>
#include<algorithm>
#include<string>

using namespace std;

typedef struct info{
	char id[7];
	char name[9];
	int grade;
}info;

bool compare1(const info &a,const info &b){
	for(int i = 0;i<6;i++){
		if(a.id[i] != b.id[i])
			return a.id[i] < b.id[i];
	}
}
bool compare2(const info &a,const info &b){
	for(int i = 0;i < 9;i++){
		if(a.name[i] != b.name[i])
			return a.name[i] < b.name[i];
	}
	for(int i = 0;i<6;i++){
		if(a.id[i] != b.id[i])
			return a.id[i] < b.id[i];
	}
}
bool compare3(const info &a,const info &b){
	if(a.grade != b.grade)
		return a.grade < b.grade;
	for(int i = 0;i<6;i++){
		if(a.id[i] != b.id[i])
			return a.id[i] < b.id[i];
	}
}

info ss[100000];

int main(void){
	int N,C;
	scanf("%d %d",&N,&C);
	for(int i = 0;i < N;i++)
		scanf("%s %s %d",ss[i].id,ss[i].name,&ss[i].grade);
	if(C==1)
		sort(ss,ss + N,compare1);
	else if(C==2)
		sort(ss,ss + N,compare2);
	else
		sort(ss,ss + N,compare3);
	for(int i = 0;i < N;i++)
		printf("%s %s %d\n",ss[i].id,ss[i].name,ss[i].grade);

	return 0;
}