MANAGER
Time Limit:1000MS    Memory Limit:10000KB    64bit IO Format:%I64d
& %I64u

Description

One of the programming paradigm in parallel processing is the producer/consumer paradigm that can be implemented using a system with a "manager" process and several "client" processes. The clients can be producers, consumers, etc.
The manager keeps a trace of client processes. Each process is identified by its cost that is a strictly positive integer in the range 1 .. 10000. The number of processes with the same cost cannot exceed 10000. The queue is managed according to three types
of requests, as follows:

  • a x - add to the queue the process with the cost x;
  • r - remove a process, if possible, from the queue according to the current manager policy;
  • p i - enforce the policy i of the manager, where i is 1 or 2. The default manager policy is 1
  • e - ends the list of requests.

There are two manager policies:

  • 1 - remove the minimum cost process
  • 2 - remove the maximum cost process

The manager will print the cost of a removed process only if the ordinal number of the removed process is in the removal list.



Your job is to write a program that simulates the manager process.

Input

The input is from the standard input. Each data set in the input has the following format:

  • the maximum cost of the processes
  • the length of the removal list
  • the removal list - the list of ordinal numbers of the removed processes that will be displayed; for example 1 4 means that the cost of the first and fourth removed processes will be displayed
  • the list of requests each on a separate line.

Each data set ends with an e request. The data sets are separated by empty lines.

Output

The program prints on standard output the cost of each process that is removed, provided that the ordinal number of the remove request is in the list and the queue is not empty at that moment. If the queue is empty the program
prints -1. The results are printed on separate lines. An empty line separates the results of different data sets.



An example is given in the following:

Sample Input

5
2
1 3
a 2
a 3
r
a 4
p 2
r
a 5
r
e

Sample Output

2
5

题目大意:

线程模拟。

ax——将一个花费为x的进程加到队列中

r——假设可能。依照当前管理者的策略,删除一个进程

p i ——运行管理者的策略i。当中i是1或者2。缺省值为1

e——请求列表终止

两个管理者的策略为:

1——删除最小耗费进程

2——删除最大耗费进程

输出指定的删除序列

#include <iostream>
#include<algorithm>
using namespace std;
int cmp1(int a,int b)
{
return a>b;
}
int cmp2(int a,int b)
{
return a<b;
}
int main()
{ int num;
while(cin>>num&&num)
{
int p=1;
int n;
cin>>n;
int a[1010]={0},b[1010]={0},c[2010]={0};
int i;
for(i=1;i<=n;i++)
cin>>b[i];
int a1=1,b1=1,c1=1;
char ch;
i=0;
while(cin>>ch&&ch!='e')
{
if(ch=='a')
{
cin>>a[a1];
a1++;
}
if(ch=='p')
cin>>p;
if(ch=='r')
{
if(p==1)//删除最小进程
{
sort(a+1,a+a1,cmp1);
c[c1]=a[a1-1];
c1++;
a1=a1-1;
}
if(p==2)//删除最大进程
{
sort(a+1,a+a1,cmp2);
c[c1]=a[a1-1];
c1++;
a1=a1-1;
}
}
}
for(i=1;i<=n;i++)
cout<<c[b[i]]<<endl;
cout<<endl;
}
return 0;
} /*
5
2
1 3
a 2
a 3
r
a 4
p 2
r
a 5
r
e
*/

刚開始提交WrongAnswer 后来注意到时sort函数的使用,数组開始下标从0開始还是从1開始sort括号中的的列表不同,

sort(a+1,a+a1,cmp1);

我的下标从1開始。

sort函数详情见http://blog.csdn.net/sunshumin/article/details/37756027

再提交时是PE错误,改成一次while循环加一个空行。ac。

POJ 1281 MANAGER的更多相关文章

  1. POJ 题目分类(转载)

    Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965 ...

  2. (转)POJ题目分类

    初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推. ...

  3. poj分类

    初期: 一.基本算法:      (1)枚举. (poj1753,poj2965)      (2)贪心(poj1328,poj2109,poj2586)      (3)递归和分治法.      ( ...

  4. poj 题目分类(1)

    poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01 ...

  5. POJ题目分类(按初级\中级\高级等分类,有助于大家根据个人情况学习)

    本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算 ...

  6. POJ题目分类(转)

    初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推. ...

  7. POJ题目细究

    acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP:  1011   NTA                 简单题  1013   Great Equipment     简单题  102 ...

  8. POJ题目(转)

    http://www.cnblogs.com/kuangbin/archive/2011/07/29/2120667.html 初期:一.基本算法:     (1)枚举. (poj1753,poj29 ...

  9. [POJ题目分类][转]

    Hint:补补基础... 初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治 ...

随机推荐

  1. XTU1201:模和除

    题目描写叙述 两个整数x和y,满足1<=x<=a,1<=y<=b 且x%y等于x/y的x和y的对数有多少? x%y是x除以y的余数, x/y是x除以y的商,即整数除. 输入 不 ...

  2. .NET 图片解密为BASE64

    #region 图片加密 /// <summary> /// 加密本地文件 /// </summary> /// <param name="inputname& ...

  3. Create a Visual C++ Wizard for Visual Studio 2005

    from:http://www.codeguru.com/cpp/v-s/devstudio_macros/customappwizards/article.php/c12775/Create-a-V ...

  4. elasticsearch如何安全重启节点

    elasticsearch如何安全重启节点 标签: elasticsearch 节点 | 发表时间:2016-05-24 03:58 | 作者:kfcman 分享到: 出处:http://www.it ...

  5. USACO 2.2 Preface Numbering

    Preface Numbering A certain book's prefaces are numbered in upper case Roman numerals. Traditional R ...

  6. Hessian Spirng实例

    Spring实例 之前,我们做了很简单的纯Hessian的调用,虽然到此已经能够满足远程调用的需求了,但是我听说spring也能够访问hessian的远程服务,研究了一番,废话不多说,直接上示例. 业 ...

  7. Ubuntu 14.04安装Skype

    Skype 4.3版本在14.04 LTS工作正常.安装步骤: $ sudo apt-get remove skype skype-bin:i386 skype:i386 $ sudo apt-get ...

  8. MySQL(十一)

    1.1 约束条件 约束条件与数据类型的宽度一样,都是可选参数 作用:用于保证数据的完整性和一致性 主要分为: PRIMARY KEY (PK) 标识该字段为该表的主键,可以唯一的标识记录 FOREIG ...

  9. Linux学习日记——字符处理

    (菜鸡学习日记 各种使用命令只列举了常用几种,发现错误望指正) 一.管道 在Linux 中,管道就是一个固定大小的缓冲区,大小为一页4K.它是一种通信的机制,可以使用管道符“ | ” 来连接进程,连接 ...

  10. Javescript 面向对象编程 — 封装

    生成实例对象的原始模式 <script type="text/javascript"> var Cat={ name:'波斯猫', color:'White' } al ...