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

结构体排序水

代码：

#include<bits/stdc++.h>
using namespace std;
struct node
{
	int x, mark;
	char name[10];
}s[1000005];
bool cmp1(node a, node b){
	return a.x < b.x;
}
bool cmp2(node a,node b){
	if(strcmp(a.name, b.name) == 0)
		return a.x < b.x;
	return  
		strcmp(a.name, b.name) < 0;
}
bool cmp3(node a, node b){
	if(a.mark != b.mark){
		return a.mark < b.mark;
	}
	return a.x < b.x;
}
int main()
{
	int n, k;
	scanf("%d%d",&n, &k);
	for(int i = 1; i <= n; i++)
		scanf("%d %s %d",&s[i].x,s[i].name,&s[i].mark);		
	if(k == 1)
		sort(s+1,s+1+n,cmp1);		
	else if(k == 2)
		sort(s+1,s+1+n,cmp2);		
	else 
		sort(s+1,s+1+n,cmp3);		
	for(int i = 1; i <= n ; i ++)
	printf("%06d %s %d\n",s[i].x,s[i].name,s[i].mark);
	return 0;
}