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 (≤10 ^​5) 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

本题就是简单的排序题，用algorithm头文件里的sort即可。最后一个测试点的超时问题：把所有的读入输出都改成printf scanf就可以过测试点了

AC代码：

#include<iostream> #include<vector> #include<algorithm> using namespace std; struct student{ string id; string name; int score; }; vector<student> list; int n,c; bool cmp1(student a,student b) { return a.id<b.id; } bool cmp2(student a,student b) { if(a.name!=b.name) return a.name<b.name; else return a.id<b.id; } bool cmp3(student a,student b) { if(a.score!=b.score) return a.score<b.score; else return a.id<b.id; } char id[10],name[10]; int grade; int main() { cin >> n >> c; for(int i=0;i<n;i++) { student a; scanf("%s%s%d",id,name,&grade); list.push_back({id,name,grade}); } if(c==1) sort(list.begin(),list.end(),cmp1); else if(c==2) sort(list.begin(),list.end(),cmp2); else sort(list.begin(),list.end(),cmp3); for(int i=0;i<list.size();i++) printf("%s %s %d\n",list[i].id.c_str(),list[i].name.c_str(),list[i].score); return 0; }