1028 List Sorting (25分)（测试点超时问题）

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 (≤10
^​5) 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

本题就是简单的排序题，用algorithm头文件里的sort即可。最后一个测试点的超时问题：把所有的读入输出都改成printf scanf就可以过测试点了

AC代码：

#include
#include
#include
using namespace std;

struct student{
     
  string id;
  string name;
  int score;
};
vector<student> list;
int n,c;

bool cmp1(student a,student b)
{
     
    return a.id<b.id;
}

bool cmp2(student a,student b)
{
     
    if(a.name!=b.name)
        return a.name<b.name;
    else
        return a.id<b.id;
}

bool cmp3(student a,student b)
{
     
    if(a.score!=b.score)
        return a.score<b.score;
    else
        return a.id<b.id;
}
char id[10],name[10];
int grade;
int main()
{
     
    cin >> n >> c;
    for(int i=0;i<n;i++)
    {
     
        student a;
        scanf("%s%s%d",id,name,&grade);
        list.push_back({
     id,name,grade});
    }
    if(c==1)
        sort(list.begin(),list.end(),cmp1);
    else if(c==2)
        sort(list.begin(),list.end(),cmp2);
    else
        sort(list.begin(),list.end(),cmp3);
    for(int i=0;i<list.size();i++)
        printf("%s %s %d\n",list[i].id.c_str(),list[i].name.c_str(),list[i].score);
    return 0;
}