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  • 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
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  • 原文地址:https://www.cnblogs.com/57one/p/11968307.html
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