A1028. List Sorting

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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

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<cstdio>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<algorithm>
 5 using namespace std;
 6 typedef struct{
 7     char id[7];
 8     char name[9];
 9     int grade;
10 }info;
11 bool cmp1(info a, info b){
12     return strcmp(a.id, b.id) < 0;
13 }
14 bool cmp2(info a, info b){
15     if(strcmp(a.name, b.name) == 0)
16         return strcmp(a.id, b.id) < 0;
17     else
18         return strcmp(a.name, b.name) < 0;
19 }
20 bool cmp3(info a, info b){
21     if(a.grade == b.grade)
22         return strcmp(a.id, b.id) < 0;
23     else
24         return a.grade < b.grade;
25 }
26 info stu[100001];
27 int main(){
28     int N, C;
29     scanf("%d%d", &N, &C);
30     for(int i = 0; i < N; i++)
31         scanf("%s %s %d", stu[i].id, stu[i].name, &stu[i].grade);
32     if(C == 1)
33         sort(stu, stu + N, cmp1);
34     else if(C == 2)
35         sort(stu, stu + N, cmp2);
36     else
37         sort(stu, stu + N, cmp3);
38     for(int i = 0; i < N; i++)
39         printf("%s %s %d\n", stu[i].id, stu[i].name, stu[i].grade);
40     cin >> N;
41     return 0;
42 }
View Code

 

总结:

1、简单题。只有一点以前没有注意到,在main函数里定义一个100000的数组时,竟然空间不足编译错误。其实只要把数组定义在main函数之外,作为全局变量即可。main函数内的变量在栈区,空间一般比较小,而全局变量一般不在栈区,空间较大。 当然也可以使用vector。

posted @ 2018-01-22 16:18  ZHUQW  阅读(154)  评论(0编辑  收藏  举报