A1028 List Sorting (25 point(s)

不同部分排序

1. 原文

原题

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

2. 解析

排序

3. AC代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
	int id,grade;
	char name[10];
}v[100010];
int c;
bool cmp(node a,node b)
{
	if (c==1)
	{
		return a.id<b.id;
	}else if(c==2){
		if (strcmp(a.name,b.name)!=0)
		{
			return strcmp(a.name,b.name)<0;
		}else{
			return a.id<b.id;
		}
	}else{
		if (a.grade!=b.grade)
		{
			return a.grade<b.grade;
		}else{
			return a.id<b.id;
		}
	}
}
int main()
{
	int n;
	scanf("%d%d",&n,&c);
	node temp;
	for (int i = 0; i < n; ++i)
	{
		scanf("%d%s%d",&temp.id,temp.name,&temp.grade);
		v[i]=temp;
	}
	sort(v,v+n,cmp);
	for (int i = 0; i < n; ++i)
	{
		printf("%06d %s %d\n",v[i].id,v[i].name,v[i].grade);
	}

	return 0;
}