B. Berland National Library
Time Limit: 2 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/567/problem/B

Description

Berland National Library has recently been built in the capital of Berland. In addition, in the library you can take any of the collected works of Berland leaders, the library has a reading room.

Today was the pilot launch of an automated reading room visitors' accounting system! The scanner of the system is installed at the entrance to the reading room. It records the events of the form "reader entered room", "reader left room". Every reader is assigned aregistration number during the registration procedure at the library — it's a unique integer from 1 to 106. Thus, the system logs events of two forms:

  • "+ ri" — the reader with registration number ri entered the room;
  • "- ri" — the reader with registration number ri left the room.

The first launch of the system was a success, it functioned for some period of time, and, at the time of its launch and at the time of its shutdown, the reading room may already have visitors.

Significant funds of the budget of Berland have been spent on the design and installation of the system. Therefore, some of the citizens of the capital now demand to explain the need for this system and the benefits that its implementation will bring. Now, the developers of the system need to urgently come up with reasons for its existence.

Help the system developers to find the minimum possible capacity of the reading room (in visitors) using the log of the system available to you.

Input

The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of records in the system log. Next follow n events from the system journal in the order in which the were made. Each event was written on a single line and looks as "+ ri" or "- ri", where ri is an integer from 1 to 106, the registration number of the visitor (that is, distinct visitors always have distinct registration numbers).

It is guaranteed that the log is not contradictory, that is, for every visitor the types of any of his two consecutive events are distinct. Before starting the system, and after stopping the room may possibly contain visitors.

Output

Print a single integer — the minimum possible capacity of the reading room.

Sample Input

6
+ 12001
- 12001
- 1
- 1200
+ 1
+ 7

Sample Output

3

HINT

题意

+ 进入的编号

-   出去的编号

最多人的时候是几个人

题解

set模拟

代码

 #include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <typeinfo>
#include <map>
#include <stack>
typedef __int64 ll;
#define inf 0x7fffffff
using namespace std;
inline ll read()
{
ll x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
} //************************************************************************************** set<int > s;
set<int >::iterator it,itt;
int main()
{ char ch;
int m;
int n=read();
int ans=;
for(int i=;i<=n;i++)
{
scanf("%c %d",&ch,&m);
if(ch=='-'){
if(s.count(m))
s.erase(m);
else
ans++;
}
else {
s.insert(m);
int j=s.size();
ans=max(j,ans);
}
getchar();
}
cout<<ans<<endl;
return ;
}

Codeforces Round #Pi (Div. 2) B. Berland National Library set的更多相关文章

  1. 构造 Codeforces Round #Pi (Div. 2) B. Berland National Library

    题目传送门 /* 题意:给出一系列读者出行的记录,+表示一个读者进入,-表示一个读者离开,可能之前已经有读者在图书馆 构造:now记录当前图书馆人数,sz记录最小的容量,in数组标记进去的读者,分情况 ...

  2. Codeforces Round #Pi (Div. 2) B. Berland National Library 模拟

    B. Berland National LibraryTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contes ...

  3. Codeforces Round #Pi (Div. 2) B Berland National Library

    B. Berland National Library time limit per test1 second memory limit per test256 megabytes inputstan ...

  4. map Codeforces Round #Pi (Div. 2) C. Geometric Progression

    题目传送门 /* 题意:问选出3个数成等比数列有多少种选法 map:c1记录是第二个数或第三个数的选法,c2表示所有数字出现的次数.别人的代码很短,思维巧妙 */ /***************** ...

  5. Codeforces Round #Pi (Div. 2)(A,B,C,D)

    A题: 题目地址:Lineland Mail #include <stdio.h> #include <math.h> #include <string.h> #i ...

  6. Codeforces Round #Pi (Div. 2) ABCDEF已更新

    A. Lineland Mail time limit per test 3 seconds memory limit per test 256 megabytes input standard in ...

  7. codeforces Round #Pi (div.2) 567ABCD

    567A Lineland Mail题意:一些城市在一个x轴上,他们之间非常喜欢写信交流.送信的费用就是两个城市之间的距离,问每个城市写一封信给其它城市所花费的最小费用和最大的费用. 没什么好说的.直 ...

  8. Codeforces Round #Pi (Div. 2) E. President and Roads tarjan+最短路

    E. President and RoadsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/567 ...

  9. Codeforces Round #298 (Div. 2) E. Berland Local Positioning System 构造

    E. Berland Local Positioning System Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.c ...

随机推荐

  1. Android Studio-设置switch/case代码块自动补齐

    相信很多和我一样的小伙伴刚从Eclipse转到Android Studio的时候,一定被快捷键给搞得头晕了,像Eclipse中代码补齐的快捷键是Alt+/ ,但是在AS中却要自己设置,这还不是问题的关 ...

  2. 深入浅出 - Android系统移植与平台开发(五)- 编译Android源码(转)

    2.3编译Android源码 Android源码体积非常庞大,由Dalvik虚拟机.Linux内核.编译系统.框架代码.Android定制C库.测试套件.系统应用程序等部分组成,在编译Android源 ...

  3. ASP.NET MVC学习笔记-----Bundles

    在网页中,我们经常需要引用大量的javascript和css文件,在加上许多javascript库都包含debug版和经过压缩的release版(比如jquery),不仅麻烦还很容易引起混乱,所以AS ...

  4. Windows 下 tail 查看日志命令工具分享

    以前在公司时服务器上面可以实现tail 命令查看程序运行日志,感觉相当不错,上网查了下这些命令是linux 下的,还好有好心人开发了一个可以在Windows下的运行的小工具,来给分享一下: 使用方法: ...

  5. thinkphp中limit方法

    limit方法也是模型类的连贯操作方法之一,主要用于指定查询和操作的数量,特别在分页查询的时候使用较多.ThinkPHP的limit方法可以兼容所有的数据库驱动类的. 用法 限制结果数量 例如获取满足 ...

  6. 没玩过这些微信小游戏你就out了

    你确定没玩过下面这些微信小游戏?是不是有点out了?赶紧添加微信号kangfuyk,回复H5马上畅玩! 当然了,扫一下二维码关注后回复H5更快捷噢! 微信小游戏列表,持续更新中 辨色大比拼!心理游戏 ...

  7. compact过滤数组中的nil

    http://ruby-doc.org/core-2.2.0/Array.html#method-i-compact compact → new_aryclick to toggle source R ...

  8. The CompilerVersion constant identifies the internal version number of the Delphi compiler.

    http://delphi.wikia.com/wiki/CompilerVersion_Constant The CompilerVersion constant identifies the in ...

  9. Rotate bitmap by real angle

    tl;dr; Use GDI+ SetWorldTransform With WinAPI's SetWorldTransform you can transform the space of dev ...

  10. HDOJ 1690

    Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total ...