1028 List Sorting

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.

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct info{
    int id;
    char name[10];
    int score;
}stu[100000];
bool cmp_1(info a, info b)
{
    return a.id < b.id;
}
bool cmp_2(info a, info b)
{
    if (strcmp(a.name, b.name) == 0) {
        return a.id < b.id;
    } else {
        return strcmp(a.name, b.name) < 0;
    }
}
bool cmp_3(info a, info b)
{
    if (a.score == b.score) {
        return a.id < b.id;
    } else {
        return a.score < b.score;
    }
}
int main()
{
    int num, mode;
    cin >> num >> mode;
    for (int i = 0; i < num; i++) {
        scanf("%d%s%d", &stu[i].id, stu[i].name, &stu[i].score);
    }
    if (mode == 1) {
        sort(stu, stu + num, cmp_1);
    } else if (mode == 2) {
        sort(stu, stu + num, cmp_2);
    } else if (mode == 3) {
        sort(stu, stu + num, cmp_3);
    }
    for (int i = 0; i < num; i++) {
        printf("%06d %s %d\n", stu[i].id, stu[i].name, stu[i].score);
    }
    return 0;
}