1028. List Sorting (25)-蒲公英云

1028. List Sorting (25)

电玩女神 2022-05-31 10:24 277阅读 0赞

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

题目大意：

Excel可以根据任何列对记录进行排序。现在你的任务是模仿这个函数。
输入：
每个输入文件包含一个测试用例。对于每个测试用例，第一行包含两个整数N(<=100000)和C，N代表记录的数量，C是应该对记录进行排序的列。然后是N行每行包含一个学生记录。一个学生记录包含他或她不同的ID（一个6位数字），姓名（一个不超过8个字符且不包含空格的字符串），和成绩（一个0到100之间的整数）。
输出：
对于每个测试用例，输出排序后的N行结果。也就是说，如果C=1，那么记录必须按照ID的增加顺序进行排序；如果C=2，则记录必须按名字的非递减顺序进行排序；如果C=3，则记录必须按照成绩的非递减顺序排序。如果有几个学生有相同的名字或成绩，他们必须按照ID的递增顺序排序。

代码：

#include<stdio.h>
#include<algorithm>
#include<vector>
#include<string.h>
using namespace std;
struct student
{
    char ID[10];
    char name[10];
    int grade;
};
vector<student> stu;
bool cmp1(student a,student b)
{
    return strcmp(a.ID,b.ID)<0;
}
bool cmp2(student a,student b)
{
    if(strcmp(a.name,b.name)!=0)
        return strcmp(a.name,b.name)<0;
    else
        return strcmp(a.ID,b.ID)<0;
}
bool cmp3(student a,student b)
{
    if(a.grade!=b.grade)
        return a.grade<b.grade;
    else
        return strcmp(a.ID,b.ID)<0;
}
int main()
{
    int i,j,n,m,k,t;
    student temp;
    scanf("%d %d",&n,&k);
    for(i=0;i<n;i++)
    {
        scanf("%s %s %d",temp.ID,temp.name,&temp.grade);
        stu.push_back(temp);
    }
    if(k==1)
    {
        sort(stu.begin(),stu.end(),cmp1);
    }
    else if(k==2)
    {
        sort(stu.begin(),stu.end(),cmp2);
    }
    else if(k==3)
    {
        sort(stu.begin(),stu.end(),cmp3);
    }
    for(i=0;i<stu.size();i++)
    {
        printf("%s %s %d\n",stu[i].ID,stu[i].name,stu[i].grade);
    }
    return 0;
}