PAT Advanced 1028 List Sorting（25）

题目描述：

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 9

Sample Output 3:

000002 James 9
000001 Zoe 60
000007 James 85
000010 Amy 90

算法描述：排序 结构体

题目大意：

给出n个同学，排序依据c（1为id 2为name 3为grade），根据c的值是1还是2还是3，对相应的列排序。第一列升序，第二列不降序，第三列不降序

#include<iostream>
#include<algorithm>
using namespace std;

struct stu
{
    string id, name;
    int grade;
}s[100010];

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

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

bool cmp3(stu a, stu b)
{
    if(a.grade == b.grade)  return a.id < b.id;
    return a.grade < b.grade;
}

int main()
{
    int n, c;
    cin >> n >> c;
    for(int i = 0 ; i < n ; i ++)
        cin >> s[i].id >> s[i].name >> s[i].grade;
    
    if(c == 1)  sort(s, s + n, cmp1);
    if(c == 2)  sort(s, s + n, cmp2);
    if(c == 3)  sort(s, s + n, cmp3);
    
    for(int i = 0 ; i < n ; i ++)
        cout << s[i].id << ' ' << s[i].name << ' ' << s[i].grade << endl;
    return 0;
}