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 (≤105) 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
解题思路:
比较简单的结构体排序,唯一需要注意的是循环中别用cin和cout,会超时。
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
struct student {
int id;
string name;
int score;
}stu[100010];
int n, c;
int cmp(student a, student b)
{
if (c == 1)
{
return a.id < b.id;
}
else if (c == 2)
{
if (a.name == b.name)
return a.id < b.id;
else return a.name < b.name;
}
else if (c == 3)
{
if (a.score == b.score)
return a.id < b.id;
else return a.score < b.score;
}
}
int main()
{
cin >> n >> c;
for (int i = 0; i < n; i++)
{
stu[i].name.resize(10);
scanf("%d %s %d", &stu[i].id, &stu[i].name[0], &stu[i].score);
}
sort(stu, stu + n, cmp);
for (int i = 0; i < n; i++)
{
printf("%06d %s %d\n", stu[i].id, stu[i].name.c_str(), stu[i].score);
}
return 0;
}