1028 List Sorting (25 分)

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 9
Sample Output 3:
000002 James 9
000001 Zoe 60
000007 James 85
000010 Amy 90

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
struct Student {
	char ID[10];
	char name[10];
	int grade;
}student[100005];
bool cmp1(Student a, Student b) {
	if (a.ID != b.ID) return strcmp(a.ID, b.ID) < 0;
}
bool cmp2(Student a, Student b) {
	if (strcmp(a.name,b.name)==0) return strcmp(a.ID, b.ID) < 0;
	else return strcmp(a.name, b.name) < 0;
}
bool cmp3(Student a, Student b) {
	if(a.grade==b.grade) return strcmp(a.ID, b.ID) < 0;
	else  return a.grade < b.grade;
}
int main() {
	int N, C;
	cin >> N >> C;
	for (int i = 0; i < N; i++) cin >> student[i].ID >> student[i].name >> student[i].grade;
	if (C == 1) {
		sort(student, student + N, cmp1);
	}
	else if (C == 2) {
		sort(student, student + N, cmp2);
	}
	else {
		sort(student, student + N, cmp3);
	}
	for (int j = 0; j < N; j++) cout << student[j].ID << " " << student[j].name << " " << student[j].grade << endl;
	return 0;
}