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

解題思路：這題沒什麼難點，只要注意資料輸入輸出用scanf，不要用cin就行，最後一個測試點會過不了，執行超時 

#include<bits/stdc++.h>
using namespace std;
struct Node{
	int id,grade;
	char name[10];
}node[100001];
int cmp1(Node a,Node b)
{
     return a.id<b.id;
}
int cmp2(Node a,Node b)
{
	if(strcmp(a.name,b.name)==0 )return a.id<b.id;
	return strcmp(a.name,b.name)<=0;
}
int cmp3(Node a,Node b)
{
	if(a.grade==b.grade) return a.id<b.id;
	return a.grade<b.grade;
}
int main(void)
{
    int n,c;
    scanf("%d %d",&n,&c);
    for(int i=0;i<n;i++)
    {
    	scanf("%d %s %d",&node[i].id,node[i].name,&node[i].grade);
    //	cin>>node[i].id>>node[i].name>>node[i].grade;
	}
	if(c==1) sort(node,node+n,cmp1);
	else if(c==2) sort(node,node+n,cmp2);
	else if(c==3) sort(node,node+n,cmp3);
	for(int i=0;i<n;i++)
	{
		printf("%06d %s %d\n",node[i].id,node[i].name,node[i].grade);
    	//cout<<" "<<node[i].name<<" "<<node[i].grade<<endl;
	}

    return 0;
}