PAT-A1028 List Sorting【排序】

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

---

 简单的结构体排序，sort自定义排序算法。虽然很简单，但是险过，算法时间和空间都很差。希望二刷可以改善这个问题

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
struct stu{
	string id;
	string name;
	int grade;
}data[100000];
bool sortById(const stu t1,const stu t2){
	return t1.id<t2.id;
}
bool sortBygrade(const stu t1,const stu t2){
	if(t1.grade<t2.grade){
		return true;
	}else if(t1.grade==t2.grade){
		return t1.id<t2.id;
	}
	return false;
}
bool sortByname(const stu t1,const stu t2){
	if(t1.name<t2.name){
		return true;
	}
	else if(t1.name==t2.name){
		return t1.id<t2.id;
	}
	return false;
}
int main(){
	int num,flag;
	string id,name;
	int grade;
	cin>>num>>flag;
	for(int a=0;a<num;a++){
		cin>>id>>name>>grade;
		data[a].id=id;
		data[a].name=name;
		data[a].grade=grade;
	}
	if(flag==1){
		sort(data,data+num,sortById);
	}
	else if(flag==2){
		sort(data,data+num,sortByname);
	}
	else if(flag==3){
		sort(data,data+num,sortBygrade);
	}
	for(int i=0;i<num;i++){
		cout<<data[i].id<<" "<<data[i].name<<" "<<data[i].grade<<endl;
	}
	return 0;
}

 一个超时，剩下的虽然过了，但是有这运行时间与空间太大的问题。针对上述代码进行优化

优化思路：

• 轻装上阵。即能少装几层有少装。比如这道题我们一拿到题就知道使用结构体保存每一个学生的数据。面对多条学生纪录，这里有两个办法进行保存，一个是使用结构体数组保存，一个是使用Vector泛型保存。使用vector相当于多套了一层，所以能不使用尽量不使用(PS,第一次代码使用的是vector也是有一次没运行超时，不过没贴代码)
• 不要脱了裤子放屁。没必要的东西，不要额外折腾。就数据的输出而言，可以直接将数据保存到数组的相应位置，而不是先保存到临时变量然后在赋值。比如36-41行代码
• 能不用高级变量的尽量不用高级变量。什么意思呢，就拿学生id来说，一共是6位数字，使用基础变量int绝对可以装得下，而我上面使用了string，姓名明确指出只有10位字符，可以使用字符数组完成，我用的string，属于额外的空间开销。
• 有现成的方法尽量不要自己自信去手写，极其不建议自己造*。比如姓名的比较，即字符串的比较，有现成得笔算算法，strcmp(char [] a,char [] b) ，返回一个int类型数字。等于0表示两个字符串相等，小于零表示第一个字符串小于第二个字符串，大于零表示第一个字符串大于第二个字符串。
• 追明显的，能用scanf输入的不用cin，能用printf输出的不用cout

---

  改良版

#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
struct Student{
    int id;
    int grade;
    char name[10];
}students[100000];
bool sortById(const Student t1,const Student t2){
	return t1.id<t2.id;
}
bool sortBygrade(const Student t1,const Student t2){
	int cmp = t1.grade - t2.grade;
        if(cmp != 0) return cmp <= 0;
        return t1.id < t2.id;
}
bool sortByname(const Student t1,const Student t2){
	
		int cmp = strcmp(t1.name, t2.name);
        if(cmp != 0) return cmp <= 0;
        return t1.id < t2.id;
}
int main(){
	int n;
	int flag;
    scanf("%d %d", &n, &flag);
    for(int i = 0; i < n; i++){
        scanf("%d %s %d", &students[i].id, students[i].name, &students[i].grade);
    }
	if(flag==1){
		sort(students, students + n, sortById);
	}else if(flag==2){
		sort(students, students + n, sortByname);
	}else if(flag==3){
		sort(students, students + n, sortBygrade);
	}
    
    for(int  i = 0; i < n; i++)
        printf("%06d %s %d\n", students[i].id, students[i].name, students[i].grade);
    return 0;
}