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  • codeforces 637B B. Chat Order(map,水题)

    题目链接:

    B. Chat Order

    time limit per test

    3 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.

    Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.

    Input

    The first line contains integer n (1 ≤ n ≤ 200 000) — the number of Polycarpus' messages. Next n lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.

    Output

    Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.

    Examples
    input
    4
    alex
    ivan
    roman
    ivan
    output
    ivan
    roman
    alex
    input
    8
    alina
    maria
    ekaterina
    darya
    darya
    ekaterina
    maria
    alina
    output
    alina
    maria
    ekaterina
    darya
    Note

    In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:

    1. alex

    Then Polycarpus writes to friend by name "ivan" and the list looks as follows:

    1. ivan
    2. alex

    Polycarpus writes the third message to friend by name "roman" and the list looks as follows:

    1. roman
    2. ivan
    3. alex

    Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:

    1. ivan
    2. roman
    3. alex

    题意:按给的顺序联系他的朋友,把最后的列表打印出来;

    思路:从后往前,没出现的输出,出现的就continue,记录是否出现过可以用map;

    AC代码:

    #include <bits/stdc++.h>
    using namespace std;
    const int N=2e5+2;
    string str[N];
    map<string,int>mp;
    int main()
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            cin>>str[i];
        }
        for(int i=n-1;i>=0;i--)
        {
            if(!mp[str[i]])
            {
                cout<<str[i]<<"
    ";
                mp[str[i]]=1;
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5280834.html
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