PAT 1028 List Sorting

Posted 2021-11-27 cunyusup

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 (≤) 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 500

struct node{
    string id;
    string name;
    int point;
};

bool comp1(node a,node b){
    return a.id < b.id;
}

bool comp2(node a,node b){
    if(a.name==b.name) return a.id < b.id;
    return a.name < b.name;
}

bool comp3(node a,node b){
    if(a.point==b.point) return a.id < b.id;
    return a.point < b.point;
}




int main(){
    int n,c;
    cin >> n >> c;
    node nod[n];
    for(int i=0;i < n;i++){
        char xid[10];
        char xname[10];
        int g;
        scanf("%s%s%d",xid,xname,&g);

//        cin >> xid >> xname >> g;
        nod[i].id = xid;
        nod[i].name = xname;
        nod[i].point = g;
    }

    if(c == 1){sort(nod,nod+n,comp1);}
    else if(c == 2){sort(nod,nod+n,comp2);}
    else if(c == 3){sort(nod,nod+n,comp3);}

    for(int i=0;i < n;i++){
//        cout << nod[i].id << " " << nod[i].name << " " << nod[i].point << endl;
        printf("%s %s %d\n",nod[i].id.c_str(),nod[i].name.c_str(),nod[i].point);
    }


    return 0;
}

char 我开成6和8就错了？？题目给的条件不对把，还是要稍微开大一点。。

当卡string输入的时候可以用char代替输入，然后再用string等于就好了，嘿嘿