• 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 }
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  • 原文地址:https://www.cnblogs.com/wydxry/p/11172786.html
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