1028 List Sorting (25 分)

1028 List Sorting (25 分)

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

三种排序方式，简单排序算是

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 struct Node
 4 {
 5     string id;
 6     string name;
 7     int sorce;
 8 }node[100005];
 9 bool cmp(Node a, Node b){
10     return a.id < b.id;
11 }
12
13 bool cmp1(Node a, Node b){
14     for(int i = 0; i < max(a.name.length(), b.name.length()); i++){
15         if(min(a.name.length(), b.name.length()) == i)
16             return a.name < b.name;
17         if(a.name[i] == b.name[i])
18             continue;
19         return a.name[i] < b.name[i];
20     }
21     return a.id < b.id;
22 }
23
24 bool cmp2(Node a, Node b){
25     if(a.sorce == b.sorce)
26         return a.id < b.id;
27     return a.sorce < b.sorce;
28 }
29
30 int n,m;
31 int main(){
32     ios::sync_with_stdio(false);
33     cin.tie(0);
34     cout.tie(0);
35     cin >> n >> m;
36     for(int i = 0 ; i < n; i++){
37         cin >> node[i].id>>node[i].name>>node[i].sorce;
38     }
39     if(m == 1){
40         sort(node,node+n,cmp);
41     }else if(m == 2){
42         sort(node,node+n,cmp1);
43     }else{
44         sort(node,node+n,cmp2);
45     }
46     for(int i = 0; i < n; i++){
47         cout <<node[i].id<<" "<<node[i].name<<" "<<node[i].sorce<<endl;
48     }
49     return 0;
50 }

原文地址：https://www.cnblogs.com/zllwxm123/p/11071164.html