题目链接:http://codeforces.com/problemset/problem/567/B

题目描述:

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 a registration 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.

Input:

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

Output:

3

Input:

2
- 1
- 2

Output:

2

题目大意:告诉你n组记录,+ id表示id进入图书馆,-为id出图书馆,求图书馆最大容量(即某时刻图书馆内的人数最多)。注意,可能在你记录前就有人在里面了。

 #include <cstdio>
#include <set>
#include <cstring>
#include <algorithm>
using namespace std; int n,num,l,mx; //l记录当前图书馆内的人数,mx维护最大容量
char ch; set<int> s; int main(){
while(~scanf("%d",&n)){
l=,mx=;
s.clear();
set<int>::iterator it;
for(int i=;i<n;i++){
getchar();
scanf("%c %d",&ch,&num);
if(ch=='+' ){
s.insert(num);
l++;
mx=max(l,mx);
}
else if(ch=='-'){
if(!s.count(num)){
mx++;
}
else{
l--;
it=s.find(num);
s.erase(it);
}
}
}
printf("%d\n",mx);
}
}

Berland National Library的更多相关文章

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

    B. Berland National LibraryTime Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest ...

  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 567B Berland National Library

    Description Berland National Library has recently been built in the capital of Berland. In addition, ...

  4. Codeforces B - Berland National Library

    B. Berland National Library time limit per test 1 second memory limit per test 256 megabytes input s ...

  5. 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 ...

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

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

  7. Codeforces 567B:Berland National Library(模拟)

    time limit per test : 1 second memory limit per test : 256 megabytes input : standard input output : ...

  8. CodeForces 567B Berland National Library hdu-5477 A Sweet Journey

    这类题一个操作增加多少,一个操作减少多少,求最少刚开始为多少,在中途不会出现负值,模拟一遍,用一个数记下最大的即可 #include<cstdio> #include<cstring ...

  9. B. Berland National Library---cf567B(set|模拟)

    题目链接:http://codeforces.com/problemset/problem/567/B  题意:题目大意: 一个计数器, +号代表一个人进入图书馆, -号代表一个人出去图书馆. 给一个 ...

随机推荐

  1. win7 安装 MongoDB 及简单操作

    下载地址 http://dl.mongodb.org/dl/win32/x86_64 这里用的版本是 mongodb-latest-signed.msi 同时下载 mongodb-compass 下载 ...

  2. setsockopt 设置socket 详细用法

    1.closesocket(一般不会立即关闭而经历TIME_WAIT的过程)后想继续重用该socket:BOOL bReuseaddr=TRUE;setsockopt(s,SOL_SOCKET ,SO ...

  3. An Introduction to Lock-Free Programming

    Lock-free programming is a challenge, not just because of the complexity of the task itself, but bec ...

  4. 第50天:scrollTo小火箭返回顶部

    scrollTo(x,y)//可把内容滚动到指定的坐标scrollTo(xpos,ypos)//x,y值必需 1.固定导航栏 <!DOCTYPE html> <html lang=& ...

  5. matlab中nargin函数的用法

    nargin是用来判断输入变量个数的函数,这样就可以针对不同的情况执行不同的功能. 通常可以用他来设定一些默认值,如下面的函数. 例子,函数test1的功能是输出a和b的和.如果只输入一个变量,则认为 ...

  6. 原生JS表单序列化

    // 表单序列化,IE9+ HTMLFormElement.prototype.serialize = function() { var form = this; // 表单数据 var arrFor ...

  7. HTML5可用的css reset

    html, body, div, span, object, iframe, h1, h2, h3, h4, h5, h6, p, blockquote, pre, abbr, address, ci ...

  8. CF1073E Segment Sum 自闭了

    CF1073E Segment Sum 题意翻译 给定\(K,L,R\),求\(L\)~\(R\)之间最多不包含超过\(K\)个数码的数的和. \(K<=10,L,R<=1e18\) 我 ...

  9. POJ. 2253 Frogger (Dijkstra )

    POJ. 2253 Frogger (Dijkstra ) 题意分析 首先给出n个点的坐标,其中第一个点的坐标为青蛙1的坐标,第二个点的坐标为青蛙2的坐标.给出的n个点,两两双向互通,求出由1到2可行 ...

  10. AOJ.865 青铜莲花池 (BFS)

    AOJ.865 青铜莲花池 (BFS) 题意分析 典型的BFS 没的说 代码总览 #include <iostream> #include <cstdio> #include ...