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;
}