題目

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 (≤ $10^{}$

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

　　使用algorithm::sort進行排序即可，可以用函式指標方便呼叫對應的排序函式。注意： 不要使用cin，否則最後一個測試點會超時。

/*
** @Brief:No.1028 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-12-04 
** @status: Accepted!
*/
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;

struct students{
	char id[10];
	char name[10];
	int grade;
};

bool cmp1(students a,students b){
	return strcmp(a.id,b.id)<=0;
}

bool cmp2(students a,students b){
	return strcmp(a.name,b.name)<=0;
}

bool cmp3(students a,students b){
	return a.grade<=b.grade;
}
int main(){
	int N,C;
	while(cin>>N>>C){
		vector<students> records;
		students input;
		for(int i=0;i<N;i++){
			scanf("%s %s %d",input.id,input.name,&input.grade);
			records.push_back(input);
		}
		bool (*cmp[])(students,students)= {cmp1,cmp2,cmp3};
		sort(records.begin(),records.end(),cmp[C-1]);
		for(auto elem:records){
			cout<<elem.id<<" "<<elem.name<<" "<<elem.grade<<endl;
		}
	}
	return 0;
}

　　cin和cout比想象中的耗時，如果使用cin和cout，會超時，如果使用scanf+cout，大約300ms，如果使用printf+scanf，只需要100ms。