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 (≤) 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

思路：

　　模拟，注意采用cin输入的话最后一组数据会被卡到。

Code：

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 struct Stu {
 6     int id;
 7     char name[10];
 8     int grade;
 9 } students[100005];
10
11 bool cmp1(Stu a, Stu b) { return a.id < b.id; }
12 bool cmp2(Stu a, Stu b) { return strcmp(a.name, b.name) <= 0; }
13 bool cmp3(Stu a, Stu b) { return a.grade <= b.grade; }
14
15 int main() {
16     int n, c;
17     scanf("%d%d", &n, &c);
18     for (int i = 0; i < n; ++i)
19         scanf("%d%s%d", &students[i].id, students[i].name, &students[i].grade);
20     if (c == 1)
21         sort(students, students + n, cmp1);
22     else if (c == 2)
23         sort(students, students + n, cmp2);
24     else
25         sort(students, students + n, cmp3);
26     for (int i = 0; i < n; ++i)
27         cout << setw(6) << setfill('0') << students[i].id << " "
28              << students[i].name << " " << students[i].grade << endl;
29     return 0;
30 }