PAT甲级——A1028 List Sorting

标签：order   incr   nes   seve   names   ons   clu   contains   any   

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

 1 #include <iostream>
 2 #include<string>
 3 #include <vector>
 4 #include<map>
 5 #include <algorithm>
 6 using namespace std;
 7 struct Node
 8 {
 9     int ID, grade;
10     string name;
11 };
12 typedef bool(*funptr)(Node a, Node b);
13 bool cmp1(Node a, Node b)
14 {
15     return a.ID < b.ID;
16 }
17 bool cmp2(Node a, Node b)
18 {
19     return (a.name == b.name) ? a.ID < b.ID : a.name < b.name;
20 }
21 bool cmp3(Node a, Node b)
22 {
23     return (a.grade == b.grade) ? a.ID < b.ID : a.grade < b.grade;
24 }
25 
26 int main()
27 {
28     int N, C;
29     cin >> N >> C;
30     vector<Node>v;
31     funptr ptr[3] = { &cmp1,&cmp2,&cmp3 };
32     for (int i = 0; i < N; ++i)
33     {
34         Node node;
35         cin >> node.ID >> node.name >> node.grade;
36         v.push_back(node);
37     }
38     //使用函数指针
39     //sort(v.begin(), v.end(), *(ptr[C - 1]));
40 
41     //使用普通方法
42     if (C == 1)
43         sort(v.begin(), v.end(), [](Node a, Node b) {return a.ID < b.ID; });
44     else if (C == 2)
45         sort(v.begin(), v.end(), [](Node a, Node b) {return (a.name == b.name) ? a.ID < b.ID : a.name < b.name; });
46     else
47         sort(v.begin(), v.end(), [](Node a, Node b) {return (a.grade == b.grade) ? a.ID < b.ID : a.grade < b.grade; });
48     for (auto a : v)
49         cout << a.ID << " " << a.name << " " << a.grade << endl;
50     return 0;
51 }

PAT甲级——A1028 List Sorting

标签：order   incr   nes   seve   names   ons   clu   contains   any   

原文地址：https://www.cnblogs.com/zzw1024/p/11219527.html