PAT甲级练习1028. List Sorting (25)

1028. List Sorting (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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

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

一道排序的题目，不难。。。但是没想到用cin，cout居然超时了。。。看来对于c++输入输出的效率这部分得去了解一下

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

struct record{
	int ID, grade;
	char name[10];
};
//按ID
bool cmp1(record r1, record r2){
	return r1.ID < r2.ID;
}
//按名字
bool cmp2(record r1, record r2){
	return strcmp(r1.name, r2.name)==0 ? r1.ID < r2.ID : strcmp(r1.name, r2.name)<0;
}
//按成绩
bool cmp3(record r1, record r2){
	return r1.grade == r2.grade ? r1.ID < r2.ID : r1.grade < r2.grade;
}
int main(){

	int n,c ;

	scanf("%d %d", &n, &c);
	vector vr;
	
	for(int i=0; i>r.ID>>r.name>>r.grade;

		vr.push_back(r);
	}

	if(c==1)sort(vr.begin(), vr.end(), cmp1);
	else if(c==2)sort(vr.begin(), vr.end(), cmp2);
	else if(c==3)sort(vr.begin(), vr.end(), cmp3);

	for(int i=0; i>n;
	return 0;
}