Memory Management

Time limit: 2.0 second
Memory limit: 64 MB

Background

Don't you know that at school pupils’ programming contest a new computer language has been developed. We call it D++. Generally speaking it doesn't matter if you know about it or not. But to run programs written in D++ we need a new operating system. It should be rather powerful and complex. It should work fast and have a lot of possibilities. But all this should be done in a future.
And now you are to… No. You should not devise the name for the operating system. You are to write the first module for this new OS. And of course it's the memory management module. Let's discuss how it is expected to work.

Problem

Our operating system is to allocate memory in pieces that we’ll call “blocks”. The blocks are to be numbered by integers from 1 up to N. When operating system needs more memory it makes a request to the memory management module. To process this request the memory management module should find free memory block with the least number. You may assume that there are enough blocks to process all requests.
Now we should define the meaning of words “free block”. At the moment of first request to the memory management module all blocks are considered to be free. Also a block becomes free when there were no requests to it during T minutes.
You may wonder about a notion “request to allocated blocks”. What does it mean, “request to allocated block”? The answer is simple: at any time the memory management module may be requested to access a given block. To process this request the memory management module should check if the requested block is really allocated. If it is, the request is considered to be successful and the block remains allocated for T minutes more. Otherwise the request is failed.
That's all about the algorithms of the memory management block. You are to implement them for N = 30 000 and T = 10 minutes.

Input

Each line of input contains a request for memory block allocation or memory block access. Memory allocation request has a form:
<Time> +
where <Time> is a nonnegative integer number not greater than 65 000. Time is given in seconds. Memory block access request has a form:
<Time> . <BlockNo>
where <Time> meets conditions mentioned above for the memory allocation request and <BlockNo> is an integer value in range from 1 to N. There will be no more than 80000 requests.

Output

For each line of input you should print exactly one line with a result of request processing. For memory allocation request you are to write an only integer — a number of allocated block. As it was mentioned above you may assume that every request can be satisfied, there will be no more than Nsimultaneously allocated blocks. For memory block access request you should print the only character:
  • '+' if request is successful (i.e. block is really allocated);
  • '-' if request fails (i.e. block with number given is free, so it can't be accessed).
Requests are arranged by their times in an increasing order. Requests with equal times should be processed as they appear in input.

Sample

input output
1 +
1 +
1 +
2 . 2
2 . 3
3 . 30000
601 . 1
601 . 2
602 . 3
602 +
602 +
1202 . 2
1
2
3
+
+
-
-
+
-
1
3
-

分析:树状数组来找当前第一个空闲内存;

   优先队列来判断内存是否过期,注意优先队列里的时间不一定是实际到期时间,要注意判断;

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=3e4+;
const int dis[][]={{,},{-,},{,-},{,}};
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p%mod;p=p*p%mod;q>>=;}return f;}
int n,m,k,t,now[maxn],a[maxn];
void add(int x,int y)
{
for(int i=x;i<=maxn-;i+=(i&(-i)))
a[i]+=y;
}
int get(int x)
{
int res=;
for(int i=x;i;i-=(i&(-i)))
res+=a[i];
return res;
}
struct node
{
int t,id;
node(int _t,int _id)
{
t=_t,id=_id;
}
bool operator<(const node&p)const
{
return t>p.t;
}
};
char op[];
priority_queue<node>q;
int main()
{
int i,j;
memset(now,-,sizeof now);
while(~scanf("%d",&n))
{
scanf("%s",op);
if(op[]=='.'){
scanf("%d",&m);
if(now[m]>=n)
{
now[m]=n+;
puts("+");
}
else puts("-");
}
else
{
while(!q.empty()&&q.top().t<n){
if(now[q.top().id]<n)add(q.top().id,-);
else q.push(node(now[q.top().id],q.top().id));
q.pop();
}
int l=,r=,ret;
while(l<=r)
{
int mid=l+r>>;
if(get(mid)<mid)ret=mid,r=mid-;
else l=mid+;
}
add(ret,);
now[ret]=n+;
q.push(node(now[ret],ret));
printf("%d\n",ret);
}
}
//system("pause");
return ;
}

ural1037 Memory Management的更多相关文章

  1. Memory Management in Open Cascade

    Open Cascade中的内存管理 Memory Management in Open Cascade eryar@163.com 一.C++中的内存管理 Memory Management in ...

  2. Java (JVM) Memory Model – Memory Management in Java

    原文地址:http://www.journaldev.com/2856/java-jvm-memory-model-memory-management-in-java Understanding JV ...

  3. Objective-C Memory Management

    Objective-C Memory Management Using Reference Counting 每一个从NSObject派生的对象都继承了对应的内存管理的行为.这些类的内部存在一个称为r ...

  4. Operating System Memory Management、Page Fault Exception、Cache Replacement Strategy Learning、LRU Algorithm

    目录 . 引言 . 页表 . 结构化内存管理 . 物理内存的管理 . SLAB分配器 . 处理器高速缓存和TLB控制 . 内存管理的概念 . 内存覆盖与内存交换 . 内存连续分配管理方式 . 内存非连 ...

  5. Android内存管理(2)HUNTING YOUR LEAKS: MEMORY MANAGEMENT IN ANDROID PART 2

    from: http://www.raizlabs.com/dev/2014/04/hunting-your-leaks-memory-management-in-android-part-2-of- ...

  6. Android内存管理(1)WRANGLING DALVIK: MEMORY MANAGEMENT IN ANDROID PART 1

    from : http://www.raizlabs.com/dev/2014/03/wrangling-dalvik-memory-management-in-android-part-1-of-2 ...

  7. Understanding Memory Management(2)

    Understanding Memory Management Memory management is the process of allocating new objects and remov ...

  8. Java Memory Management(1)

    Java Memory Management, with its built-in garbage collection, is one of the language’s finest achiev ...

  9. 再谈.net的堆和栈---.NET Memory Management Basics

    .NET Memory Management Basics .NET memory management is designed so that the programmer is freed fro ...

随机推荐

  1. Java中精确的数字计算类BigDecimal

    在日常开放当中需要我们计算数字,利率.通常Java的做法是使用Math相关的API.但是,这样做是不够精确的,由于float和double不能进行计算,如果强行进行计算会使得计算不准确.造成难以挽回的 ...

  2. www.iis.net

    http://www.iis.net 这是一个神奇的网站 关于IIS的所有管理,在这里都能找到 今天,一个同事问我,  iis8 php的设置,一个环境变量的东西不知道怎么去设置, 然后我搜了下,在 ...

  3. Windows如何压缩tar.gz格式

    Windows如何压缩tar.gz格式 tar.gz 是linux和unix下面比较常用的格式,几个命令就可以把文件压缩打包成tar.gz格式 然而这种格式在windows并不多见,WinRAR.Wi ...

  4. DOM操作-动态创建网页元素

    动态创建新的DOM元素,是JavaScript操作网页对象模型的重要手段之一 代码: <!DOCTYPE html> <html> <head> <title ...

  5. wamp服务器

    wamp2.5会出现的问题,是因为WampServer2.5内置的是PHP5.5,而PHP5.5 环境是基于VC11的编译脚本下生成的,所以需要安装Visual Studio 2012 VC 11 v ...

  6. openwrt 中make的使用

    make 命令简单说明 make V=99 V=99表示输出详细的debug信息 make world 表示编译所有 make j=2 V=99 如是多核CPU,加j=2 选项理论上能加快编译速度 m ...

  7. flash 右键菜单隐藏与修改

    来源:http://blog.sina.com.cn/s/blog_7264c84401014fmd.html import flash.ui.ContextMenu;import flash.ui. ...

  8. 搭建Ubuntu下c/c++编译环境【转】

    1.       安装Ubuntu. 2.       安装gcc 方法一: sudo apt-get  install  build-essential 安装完了可以执行 gcc--version的 ...

  9. 状压dp找寻环的个数 Codeforces Beta Round #11 D

    http://codeforces.com/problemset/problem/11/D 题目大意:给你n个点,m条边,找该图中有几个换 思路:定义dp[i][j]表示i是圈的集合,j表示该集合的终 ...

  10. DotNetBar教程

    DotNetBar是一组用于.NET Framework环境下的一组组件集,利用该组件集能够打造绚丽并且实用的应用程序界面,给开发人员提供极大的便利.关于DotNetBar,详情请参考其官方网站:ht ...