PAT 1028 List Sorting (25)

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本文介绍了一种模拟Excel排序功能的方法，通过C++实现对学生记录按ID、姓名或成绩排序的功能，并分享了从使用结构体到优化比较函数以解决超时问题的经验。

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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

解答：该题我刚开始直接用结构体记录，然后用使用STL库sort和stable_sort进行排序，后来最后一种情况出现超时的情况。后来将cin和cout都改成了scanf和printf，但是仍出现超时的情况。再后来将id的比较用int来换string，才解决了超时的问题。

AC代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>

using namespace std;

typedef struct{
	int id;
	char name[10];
	int score;
}record;

bool sort_id(record r1, record r2)
{
	return r1.id < r2.id;
}

bool sort_name(const record r1, const record r2)
{
	return strcmp(r1.name, r2.name);
}

bool sort_score(const record r1, const record r2)
{
	return r1.score < r2.score;
}

int main()
{
	int N, C;
	record row;
	vector<record> records;
	
	scanf("%d %d", &N, &C);
	while(N--)
	{
		//cin >> row.id >> row.name >> row.score;  cin 和cout 速度慢 
		scanf("%d%s%d", &row.id, row.name, &row.score);
		records.push_back(row);
	}
	switch(C)
	{
		case 1:
			sort(records.begin(), records.end(), sort_id);
			break;
		case 2:
			sort(records.begin(), records.end(), sort_id);
			stable_sort(records.begin(), records.end(), sort_name);
			break;
		case 3:
			sort(records.begin(), records.end(), sort_id);
			stable_sort(records.begin(), records.end(), sort_score);
			break;
	}
	for(int i = 0; i < records.size(); ++i)
	{
		printf("%06d %s %d\n",records[i].id, records[i].name, records[i].score); 
	}
	return 0;
}

当然，也可以将id当成char[]来处理，只不过，在比较大小时，需要先进行hash下。

AC代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>

using namespace std;

typedef struct{
	char id[10];
	char name[10];
	int score;
}record;

int Hash(const char id[])
{
	return (id[0] - '0')*100000 
			+ (id[1] - '0')*10000 
			+ (id[2] - '0')*1000 
			+ (id[3] - '0')*100
			+ (id[4] - '0')*10
			+ (id[5] - '0'); 
}
bool sort_id(const record r1, const record r2)
{
	return Hash(r1.id) < Hash(r2.id);
}

bool sort_name(const record r1, const record r2)
{
	return strcmp(r1.name, r2.name);
}

bool sort_score(const record r1, const record r2)
{
	return r1.score < r2.score;
}

int main()
{
	int N, C;
	record row;
	vector<record> records;
	
	scanf("%d %d", &N, &C);
	while(N--)
	{
		//cin >> row.id >> row.name >> row.score;  cin 和cout 速度慢 
		scanf("%s%s%d", row.id, row.name, &row.score);
		records.push_back(row);
	}
	switch(C)
	{
		case 1:
			sort(records.begin(), records.end(), sort_id);
			break;
		case 2:
			sort(records.begin(), records.end(), sort_id);
			stable_sort(records.begin(), records.end(), sort_name);
			break;
		case 3:
			sort(records.begin(), records.end(), sort_id);
			stable_sort(records.begin(), records.end(), sort_score);
			break;
	}
	for(int i = 0; i < records.size(); ++i)
	{
		printf("%s %s %d\n",records[i].id, records[i].name, records[i].score); 
	}
	return 0;
}