A1028

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

#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
struct student{
    string name;
    int id,score;
}p[100005];
int cmp1(struct student x,struct student y){
    return x.id<y.id;
}
int cmp2(struct student x,struct student y){
    if(x.name!=y.name)return x.name<y.name;
    else return x.id<y.id;
}
int cmp3(struct student x,struct student y){
    if(x.score!=y.score)return x.score<y.score;
    else return x.id<y.id;
}
int main(){
    int N,C,id,score;
    string name;
    cin>>N>>C;
    for(int i=0;i<N;i++){
        scanf("%d %s %d",&id,name.c_str(),&score);
        p[i].id=id;p[i].name=name.c_str();p[i].score=score;
    }
    if(C==1)sort(p,p+N,cmp1);
    if(C==2)sort(p,p+N,cmp2);
    if(C==3)sort(p,p+N,cmp3);
    for(int i=0;i<N;i++){
        printf("%06d %s %d\n",p[i].id,p[i].name.c_str(),p[i].score);
    }
    return 0;
}