PAT 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

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<cstring>
 5 using namespace std;
 6 struct node{
 7   char name[9];
 8   int id, score;
 9 };
10 
11 int c;
12 bool cmp(node a, node b){
13   if(c==1) return a.id<b.id;
14   if(c==2) return strcmp(a.name, b.name)==0 ? a.id<b.id : strcmp(a.name, b.name)<0;
15   if(c==3) return a.score==b.score? a.id<b.id : a.score<b.score;
16 }
17 int main(){
18   int n, i;
19   scanf("%d%d", &n, &c);
20   vector<node> v(n);
21   for(i=0; i<n; i++) scanf("%d %s %d", &v[i].id, v[i].name, &v[i].score);
22   sort(v.begin(), v.end(), cmp);
23   for(i=0; i<n; i++) printf("%06d %s %d
", v[i].id, v[i].name, v[i].score);
24   return 0;
25 }
原文地址:https://www.cnblogs.com/mr-stn/p/9545120.html