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("d %s %d\n", v[i].id, v[i].name, v[i].score); 24 return 0; 25 }

 

转载于:https://www.cnblogs.com/mr-stn/p/9545120.html