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. IE浏览器模式设置

    文件兼容性用于定义让IE如何编译你的网页.此文件解释文件兼容性,如何指定你网站的文件兼容性模式以及如何判断一个网页该使用的文件模式. 前言 为了帮助确保你的网页在所有未来的IE版本都有一致的外观,IE ...

  2. c# 重写索引

    //using System;//using System.Collections.Generic;//using System.Text; //namespace 索引//{//    class ...

  3. Matlab之字符串处理

    Matlab处理字符串 1.取得部分字符串 我们有一个字符串 file='20131030_113109.TemporaryAlias.Poly5'; 简单操作举例: >> a=file( ...

  4. Emag eht htiw Em Pleh(imitate)

    Emag eht htiw Em Pleh Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2901   Accepted:  ...

  5. Java中的异常

    一.什么是异常 异常就是在程序的运行过程中所发生的不正常的事件,如所需文件找不到,网络连接不通或中断,算术运算出错(如被0除),数组下标越界,装载了一个不存在的类,对null的操作,类型转换异常等等. ...

  6. Linux解压安装与卸载

    linux tar.gz zip 解压缩 压缩命令 linux下安装软件主要有这么几种: 1.自动安装: yum install package 2.用二进制文件安装:rpm -ivh file.rp ...

  7. poj2485 Highways

    Description The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public h ...

  8. Linux Apache和Nginx网络模型详解

    进程阻塞和挂起的定义: 阻塞是由于进程所需资源得不到满足,并会最终导致进程被挂起     进程挂起的原因并不一定是由于阻塞,也有可能是时间片得不到满足,挂起状态是进程从内存调度到外存中的一种状态,若在 ...

  9. Java正则表达式的最简单应用

    String emailRegex = "^\\w+([-+.]\\w+)*@\\w+([-.]\\w+)*\\.\\w+([-.]\\w+)*$"; Pattern pat = ...

  10. snoopy 强大的PHP采集类使用实例代码

    下载地址: http://www.jb51.net/codes/33397.html Snoopy的一些特点: 1抓取网页的内容 fetch 2 抓取网页的文本内容 (去除HTML标签) fetcht ...