PAT (Advanced Level) Practice 1028 List Sorting (25 分) (自定义排序)

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

自定义结构体排序
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <string>
 4 #include <cstring>
 5 using namespace std;
 6 struct node
 7 {
 8     string str1,str2;
 9     int str3;
10 }a[100005];
11 int cmp1(node x,node y)
12 {
13     return x.str1<y.str1;
14 }
15 int cmp2(node x,node y)
16 {
17     if(x.str2==y.str2) return x.str1<y.str1;
18     return x.str2<y.str2;
19 }
20 int cmp3(node x,node y)
21 {
22     if(x.str3==y.str3) return x.str1<y.str1;
23     return x.str3<y.str3;
24 }
25 int main()
26 {
27     int n,m;
28     while(cin>>n>>m){
29         for(int i=0;i<n;i++){
30             cin>>a[i].str1>>a[i].str2>>a[i].str3;
31         }
32         if(m==1){
33             sort(a,a+n,cmp1);
34         }else if(m==2){
35             sort(a,a+n,cmp2);
36         }else if(m==3){
37             sort(a,a+n,cmp3);
38         }
39         for(int i=0;i<n;i++){
40             cout<<a[i].str1<<" "<<a[i].str2<<" "<<a[i].str3<<endl;
41         }
42     }
43     return 0;
44 }

 

posted @ 2019-07-11 21:11  wydxry  阅读(201)  评论(0编辑  收藏  举报
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