PTA(Advanced Level)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

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

思路

• 仍旧是结构体数组排序的任务，根据C取值的不同方式进行排序，对应有三个排序函数cmp

代码

#include<bits/stdc++.h>
using namespace std;
struct student
{
	char id[7];
	char name[9];
	int grade;
}a[100010];

bool cmp1(student a, student b)
{
	return strcmp(a.id, b.id) < 0;
}   //比较id

bool cmp2(student a, student b)
{
	if(strcmp(a.name, b.name) != 0)
		return strcmp(a.name, b.name) < 0;
	else
		return strcmp(a.id, b.id) < 0;
}//比较名字，相同就比较id

bool cmp3(student a, student b)
{
	if(a.grade != b.grade)
		return a.grade < b.grade;
	else
		return strcmp(a.id, b.id) < 0;
}//比较分数，相同就比较id

int main()
{
	int n, c;
	cin >> n >> c;
	for(int i=0;i<n;i++)
	{
		cin >> a[i].id >> a[i].name >> a[i].grade;
	}
	switch(c)
	{
		case 1:
			sort(a, a+n, cmp1);
			break;  //执行完一个分支要break
		case 2:
			sort(a, a+n, cmp2);
			break;  //执行完一个分支要break
		case 3:
			sort(a, a+n, cmp3);
	}
	for(int i=0;i<n;i++)
	{
		cout << a[i].id << " " << a[i].name << " " <<a[i].grade;
		if(i != n-1)    cout << endl;   //避免输出额外空格
	}
	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805468327690240