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
分析:就是排序的一道题 要注意不能使用cin和cout 不然会超时
用 cin和 cout写法
 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include<iostream>
 3 #include<string>
 4 #include<stdlib.h>
 5 #include<vector>
 6 #include<algorithm>
 7 
 8 using namespace std;
 9 typedef string s[3];
10 int N, C;
11 bool compare(const vector<string> &s1,const vector<string> &s2)
12 {
13     return (s1[C - 1] != s2[C - 1]) ? s1[C - 1] < s2[C - 1] : s1[0] < s2[0];
14 }
15 int main()
16 {
17     cin >> N >> C;
18     vector<vector<string> > T(N);
19     s str;
20     for (int i = 0; i < N; i++)
21     {
22         cin >> str[0] >> str[1] >> str[2];
23         T[i].push_back(str[0]);
24         T[i].push_back(str[1]);
25         T[i].push_back(str[2]);
26     }
27     sort(T.begin(), T.end(), compare);
28     for (auto it : T)
29         cout << it[0] <<" "<< it[1]<<" "<< it[2] << endl;
30 }
View Code
原文地址:https://www.cnblogs.com/57one/p/11968307.html