1028 List Sorting

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

---

题型分类：排序

题目大意：给出学生的id、姓名和成绩，对选中的列进行排序

解题思路：用switch来控制对哪一列进行排序

---

#include 
#include 
#include 

using namespace std;

const int maxn = 100010;

typedef struct {
	int id;
	char name[10];
	int grade;
}student;

student stu[maxn];

bool cmpId(student a, student b);
bool cmpName(student a, student b);
bool cmpGrade(student a, student b);

int main(int argc, char** argv) {
	int N, C;
	scanf("%d %d", &N, &C);
	for(int i = 0; i < N; i++){
		scanf("%d %s %d", &stu[i].id, stu[i].name, &stu[i].grade);
	}
	switch(C){
		case 1:{
			sort(stu, stu + N, cmpId);
			break;
		}
		case 2:{
			sort(stu, stu + N, cmpName);
			break;
		}
		case 3:{
			sort(stu, stu + N, cmpGrade);
			break;
		}
	}
	for(int i = 0; i < N; i++){
		printf("%06d %s %d\n", stu[i].id, stu[i].name, stu[i].grade);
	}
	return 0;
}

bool cmpId(student a, student b){
	return a.id < b.id;
}

bool cmpName(student a, student b){
	if(strcmp(a.name, b.name) != 0) return strcmp(a.name, b.name) < 0;
	else return a.id < b.id;
}

bool cmpGrade(student a, student b){
	if(a.grade != b.grade) return a.grade < b.grade;
	else return a.id < b.id;
}