甲级PAT 1028 List Sorting（sort排序）

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本文介绍了一个排序算法，用于根据ID、姓名或成绩对学生记录进行排序。通过定义比较函数并使用sort()函数，可以实现不同标准下的排序，确保了数据的有序性和准确性。

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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 (≤105) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then Nlines 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

题目要求：

给出N个学生的ID，姓名以及成绩，按照三种方式排序。

C==1时，按照ID递增的顺序排序，C==2时按照姓名非递减的顺序排序，C==3按照成绩非递减的顺序排序。

若学生姓名和成绩相同，则按照学号递增的顺序排序

解题思路：

写三个comp() 然后用sort()排序。

完整代码：

#include<bits/stdc++.h>
using namespace std;

struct Student{
	int id;
	string name;
	int score;
}stu[100010]; 

bool comp1(Student stu1,Student stu2){
	return stu1.id < stu2.id;
}

bool comp2(Student stu1,Student stu2){
	if(stu1.name.compare(stu2.name)<0) return true;
	else if(stu1.name.compare(stu2.name)==0){
		if(stu1.id < stu2.id) return true;
	}
	return false;
}

bool comp3(Student stu1,Student stu2){
	if(stu1.score < stu2.score) return true;
	else if(stu1.score == stu2.score){
		if(stu1.id < stu2.id) return true;
	}
	return false;
}

int main(){
	int C,N,i;
	cin>>N>>C;
	for(i=0;i<N;i++){
		cin>>stu[i].id>>stu[i].name>>stu[i].score;
	}
	switch(C){
		case 1:sort(stu,stu+N,comp1);break;
		case 2:sort(stu,stu+N,comp2);break;
		case 3:sort(stu,stu+N,comp3);break;
	}
	for(i=0;i<N;i++){
		printf("%06d ",stu[i].id);
		cout<<stu[i].name<<" "<<stu[i].score<<endl;
	}
	return 0;
}

代码简洁

2019.2.8

#include<bits/stdc++.h>
using namespace std;

struct Rec{
	int id,grade;
	char name[10];
};
Rec rec[100010];
bool com1(Rec r1,Rec r2){
	return r1.id<r2.id;
}
bool com2(Rec r1,Rec r2){
	if(strcmp(r1.name,r2.name)<0) return true;
	else if(strcmp(r1.name,r2.name)==0) return r1.id < r2.id;
	else return false;
}
bool com3(Rec r1,Rec r2){
	if(r1.grade!=r2.grade) return r1.grade < r2.grade;
	else return r1.id < r2.id;
}

int main(){
	int i,n,c;
	scanf("%d %d",&n,&c);
	for(i=0;i<n;i++) scanf("%d %s %d",&rec[i].id,rec[i].name,&rec[i].grade);
	switch(c){
		case 1:sort(rec,rec+n,com1);break;
		case 2:sort(rec,rec+n,com2);break;
		case 3:sort(rec,rec+n,com3);break;
	}
	for(i=0;i<n;i++) printf("%06d %s %d\n",rec[i].id,rec[i].name,rec[i].grade); 
	return 0;
}