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

 #include <iostream>
 #include <algorithm>
 #include <string>
 #include <cstring>
 using namespace std;
 struct node
 {
     string str1,str2;
     int str3;
 }a[];
 int cmp1(node x,node y)
 {
     return x.str1<y.str1;
 }
 int cmp2(node x,node y)
 {
     if(x.str2==y.str2) return x.str1<y.str1;
     return x.str2<y.str2;
 }
 int cmp3(node x,node y)
 {
     if(x.str3==y.str3) return x.str1<y.str1;
     return x.str3<y.str3;
 }
 int main()
 {
     int n,m;
     while(cin>>n>>m){
         for(int i=;i<n;i++){
             cin>>a[i].str1>>a[i].str2>>a[i].str3;
         }
         if(m==){
             sort(a,a+n,cmp1);
         }else if(m==){
             sort(a,a+n,cmp2);
         }else if(m==){
             sort(a,a+n,cmp3);
         }
         for(int i=;i<n;i++){
             cout<<a[i].str1<<" "<<a[i].str2<<" "<<a[i].str3<<endl;
         }
     }
     return ;
 }