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 (≤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 9

Sample Output 3:

000002 James 9
000001 Zoe 60
000007 James 85
000010 Amy 90

思路

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Stu{
	char id[10];
	char name[10];
	int grade;
}stu[100010];

bool cmp1(Stu a,Stu b){
	return strcmp(a.id,b.id) < 0;
}
bool cmp2(Stu a,Stu b){
	if(strcmp(a.name,b.name) != 0) return strcmp(a.name,b.name) < 0;
	else return strcmp(a.id,b.id) < 0;

}
bool cmp3(Stu a,Stu b){
	if(a.grade != b.grade) return a.grade < b.grade;
	else return strcmp(a.id,b.id) < 0;
}

int main(){
	int n,c;
	scanf("%d%d",&n,&c);
	for(int i = 0; i < n; i++){
		scanf("%s %s %d", &stu[i].id,&stu[i].name,&stu[i].grade);
	}
	if(c == 1) sort(stu,stu+n,cmp1);
	else if(c==2) sort(stu,stu+n,cmp2);
	else sort(stu,stu+n,cmp3);	
	for(int i = 0; i < n; i++){
		printf("%s %s %d\n",stu[i].id,stu[i].name,stu[i].grade);
	}
	return 0;
	
}