VK Cup 2016 - Qualification Round 1 (Russian-Speaking Only, for VK Cup teams) B. Chat Order 水题

B. Chat Order

题目连接：

http://www.codeforces.com/contest/637/problem/B

Description

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.

Sample Input

4

alex

ivan

roman

ivan

Sample Output

ivan

roman

alex

Hint

题意

有一个人，有一个队列。

然后现在依次插入n个单词，如果这个单词已经在队列中了，那就把这个单词从原来位置扔到队列首。

否则就把这个单词扔到队首。

问你最后这个队列长啥样。

题解：

其实倒着输出就好了，然后开个map记录一下vis就行了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
string s[maxn];
map<string,int>vis;
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
        cin>>s[i];
    for(int i=n;i;i--)
    {
        if(vis[s[i]])continue;
        cout<<s[i]<<endl;
        vis[s[i]]=1;
    }
}