【PAT】甲级1028 List Sorting (25 分)

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

#include
using namespace std;

int n,c;  // 总共输入三行

// 定义学生信息结构体
struct NODE
{
    int no;
    string name;
    int score;
};

int cmp(NODE a, NODE b)
{
    if(c == 1)
    {
        return a.no < b.no;// 从小到大顺序
    }
    else if(c == 2)
    {   // If there are several students who have the same name or grade, they must be sorted according to their ID's in increasing order.
        if(a.name != b.name)
        {
            return a.name < b.name;
        }
        else{
            return a.no < b.no;
        }
    }
    else if(c == 3){
        if(a.score != b.score)
        {
            return a.score < b.score;
        }
        else{
            return a.no < b.no;
        }
    }
}

int main()
{
    cin>>n>>c;
    vector node(n); // 定义一个数组存储三个学生的结构体

    // 读入学生信息
    for(int i = 0; i < n; i++)
    {
        cin>>node[i].no>>node[i].name>>node[i].score;
    }

    // 排序  注意第三个参数：定义排序的规则
    sort(node.begin(),node.end(),cmp);

    for(int i = 0; i < n; i++)
    {
        printf("%06d %s %dn",node[i].no,node[i].name.c_str(),node[i].score);
    }

    return 0;
}