PAT 甲级 1028 List Sorting (25 分) STL +sort-蒲公英云

PAT 甲级 1028 List Sorting (25 分) STL +sort

灰太狼 2024-02-19 21:29 156阅读 0赞

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 (≤105) 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

注意排序规则

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node{
    string id,name;
    int grade;
};
int cmp1(node a,node b)
{
    return a.id<b.id;
}
int cmp2(node a,node b)
{
    if(a.name==b.name)
    return a.id<b.id;
    else
    return a.name<b.name;
}
int cmp3(node a,node b)
{
    if(a.grade==b.grade)
    return a.id<b.id;
    else
    return a.grade<b.grade;
}
int main()
{
    int n,c,i,j;
    node t;
    vector<node> s;
    cin>>n>>c;
    for(i=0;i<n;i++)
    {
        cin>>t.id>>t.name>>t.grade;
        s.push_back(t);
    }
    switch(c)
    {
        case 1:sort(s.begin(),s.end(),cmp1);break;
        case 2:sort(s.begin(),s.end(),cmp2);break;
        case 3:sort(s.begin(),s.end(),cmp3);break;
        default: break;
    }
    for(i=0;i<s.size();i++)
    cout<<s[i].id<<" "<<s[i].name<<" "<<s[i].grade<<endl;
    return 0;
}