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 (≤) 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

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

struct student{
    int id;
    char name[10];
    int grade;
}stu[100010];

bool cmp1(student a, student b)
{//第一个排序函数

    return a.id < b.id;
}

bool cmp2(student a, student b)
{
    int s = strcmp(a.name , b.name);
   if(s!=0)//只要名字不相同
        return s<0;
   else
        return  a.id < b.id;
}

bool cmp3(student a, student b)
{
    if(a.grade != b.grade)
        return a.grade < b.grade;
    else
        return a.id < b.id;
}

int main()
{

    int n;
    int c;
    scanf("%d %d",&n,&c);
    for(int i=0;i<n;i++)
    {
        scanf("%d %s %d",&stu[i].id,stu[i].name,&stu[i].grade);
    }
    if(c==1)//按照c的值来排序
        sort(stu,stu+n,cmp1);
    else if(c==2)
        sort(stu,stu+n,cmp2);
    else if(c==3)
        sort(stu,stu+n,cmp3);

    for(int i=0;i<n;i++)
        printf("%06d %s %d\n",stu[i].id,stu[i].name,stu[i].grade);

    return 0;
}