【PAT甲级】 List Sorting

喜欢ctrl的cxk

发布于 2019-11-08 13:50:58

2630

发布于 2019-11-08 13:50:58

文章被收录于专栏：Don的成长史

本文链接：https://blog.csdn.net/weixin_42449444/article/details/88828816

Problem Description：

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

解题思路：

这道水题其实并不难，主要就是我这个英语水平读题的时候比较吃力（心中暗暗立下flag：一定要好好恶补英语）。根据C的取值来分情况进行排序，最后无脑for-each进行输出即可。可惜提交代码之后出现了万恶的TLE啊，25分的题只得了21！然后我取消cin和stdin的同步之后还是TLE，那没得一点办法我只能用printf代替cout进行输出以减少一点点运行时间。诶嘿提交之后AC啦！

AC代码：

#include <bits/stdc++.h>
using namespace std;

struct stu
{
    string id,name;   //学生的id、姓名
    int grade;   //学生的成绩
};

bool cmp1(stu a,stu b)  //根据id递增排序
{
    return a.id < b.id;
}

bool cmp2(stu a,stu b)  //根据姓名非递减排序,当姓名相同时按照id递增排序
{
    return a.name!=b.name ? a.name<b.name : a.id<b.id;
}

bool cmp3(stu a,stu b)  //根据成绩非递减排序,当成绩相同时按照id递增排序
{
    return a.grade!=b.grade ? a.grade<b.grade : a.id<b.id;
}

int main()
{
    ios::sync_with_stdio(false);    //取消cin和stdin的同步之后还是会TLE
    int N,C;
    cin >> N >> C;
    vector<stu> v;
    for(int i = 0; i < N; i++)
    {
        string id,name;
        int grade;
        cin >> id >> name >> grade;
        v.push_back({id,name,grade});
    }
    switch(C)
    {
        case 1: sort(v.begin(),v.end(),cmp1); break;
        case 2: sort(v.begin(),v.end(),cmp2); break;
        case 3: sort(v.begin(),v.end(),cmp3); break;
        default: break;
    }
    for(auto it : v)
    {
        //cout << it.id << " " << it.name << " " << it.grade << endl;
        //取消cin和stdin的同步之后还是会超时,必须用printf来代替cout进行输出
        printf("%s %s %d\n",it.id.c_str(),it.name.c_str(),it.grade);
    }
    return 0;
}

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2019/03/26 ，如有侵权请联系 cloudcommunity@tencent.com 删除

linux

本文分享自作者个人站点/博客前往查看

如有侵权，请联系 cloudcommunity@tencent.com 删除。

本文参与腾讯云自媒体同步曝光计划，欢迎热爱写作的你一起参与！

linux

登录后参与评论

0 条评论

热度