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 90

Sample Output 3:

000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90
#include 
#include 
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn= 100000+5;

struct Node{
	int id;
	char name[10];
	int score;
	
	
}stu[maxn];

bool cmp1(Node a,Node b){
	 return a.id-b.id<0;
}

bool cmp2(Node a,Node b){
	 if( strcmp(a.name,b.name)!=0)return  strcmp(a.name,b.name)<0; 
		else  return	a.id-b.id<0;
}
bool cmp3(Node a,Node b){
	 if( a.score!=b.score)return  a.score>n>>k;
	for(int i=0;i>stu[i].id>>stu[i].name>>stu[i].score;	
	}
	if(k==1) {
		
		sort(stu,stu+n,cmp1);	
	}
	if(k==2) {
		
		sort(stu,stu+n,cmp2);	
	}
	if(k==3) {
		sort(stu,stu+n,cmp3);	
	}
	for(int i=0;i

 

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