1028 List Sorting (25 分)

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

 Their own practices:

#include <iostream>
#include <vector>
#include <algorithm>	
#include <string>	

using namespace std;

struct record
{
	string id, name;
	int grade;
};

bool cmp1(record a, record b)
{
	return a.id < b.id;
}

bool cmp2(record a, record b)
{
	return a.name != b.name ? a.name < b.name : a.id < b.id;
}

bool cmp3(record a, record b)
{
	return a.grade != b.grade ? a.grade < b.grade : a.id < b.id;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	int n, c;
	cin >> n >> c;
	vector<record> r(n);
	for (int i = 0; i < n; i++)
	{
		cin >> r[i].id >> r[i].name >> r[i].grade;
	}
	switch (c)
	{
    default:    //没有这个default  最后一个测试点会不过
        break;
	case 1:
		sort(r.begin(),r.end() , cmp1);
		break;
	case 2:
		sort(r.begin(), r.end(), cmp2);
		break;
	case 3:
		sort(r.begin(), r.end(), cmp3);
		break;
	}
	for (int i = 0; i < n; i++)
	{
		cout << r[i].id << " " << r[i].name << " " << r[i].grade << endl;
	}
	//fclose(stdin);
	return 0;
}