Cellular Network

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

A cellular network is a radio network made up of a number of cells each served by a base station located in the cell. The base station receives call signals from mobile users (mobiles) in the cell it serves, which then connects the calls to the wired land-line telephone network. When a call is requested to connect to a mobile, the cellular network must know in which cell the mobile is located so that the call is routed to the base station of the cell appropriately.

Mobiles move from once cell to another in a cellular network. Whenever a mobile reports its new cell as it crosses boundaries of cells, the cellular network would know its exact cell at any time and finding (paging) the mobile becomes a trivial task. But it is usually infeasible for a mobile to report its new location each time it enters a new cell because of the insufficiencies of resources such as the radio bandwidth. But normally at the time of a call arrival, the cellular network knows a limited number of cells where the mobile is located. In this situation, a lot of paging strategies are developed to locate a mobile efficiently. The ultimate goal of paging strategies is to minimize both the time delay and cost of paging until the mobile is found.

Now we define our problem formally. The location area is the set of n cells C = {c1c2,..., cn} such that the mobile is guaranteed to be in one of these cells at the time of a call arrival. Suppose that it is possible to page any subset of these n cells in a unit of time (paging rounds) and find out if the mobile is located in one of the cells paged. The fastest strategy to find the cell where the mobile is located is to page all the n cells in the first and only round. However this strategy uses a lot of wireless bandwidth.

In many cases, the cellular network knows about the whereabouts of the mobile. This knowledge can be modeled with n probability values, where the probability of the mobile being present in a cell can be estimated for each of these n cells at the time of a call arrival. Let pi be the probability that the mobile is located at the cell ci and all the probabilities are independent. A sequential paging strategy is to page the cells sequentially in n paging rounds terminating once the mobile is found. Then the average cost of paging (number of cells paged),  , and the average paging delay (number of paging rounds) in locating the mobile,, can be expressed as follows:

(i x pi),(i x pi).

The parallel paging strategy is to page the cells in a collection of cells simultaneously. Sequential paging strategy has lower paging cost than parallel paging strategy, but at the expense of larger paging delay. The method of parallel paging is to partition the cells in a location area into a series of indexed groups referred to as paging zones. Let Z1Z2,..., Zw be the partition of the location area C (i.e., a partition of C into w groups), where each Zi is non-empty and corresponds to a distinct paging zone. When a call arrives for a mobile, the cells in the first paging zone Z1 are paged simultaneously in the first round and then if the mobile is not found in the first round of paging, all the cells in the second paging zone Z2 are paged, and so on. Let the number of cells in the paging zone Zi be denoted by ni = | Zi|, and let  be the corresponding zone probabilities of the users in the paging zone Zi, where  = pj. Then the average cost of paging (number of cells paged), , and the average paging delay (number of paging rounds) in locating the mobile, D, can be expressed as follows:

 = (),(i x ).

In parallel paging strategy, there is a tradeoff between bandwidth for time. For example, we increases the number of paging zones, then the paging cost could be decreased. If we decrease the number of paging zones, then the paging cost could be increased. Furthermore, for a fixed number w of paging zones, the paging cost could be different to the strategies how the cells in location area are partitioned.

For example, there are n = 5 cells in a location area C = {c1c2,..., c5} and the probability of each cells in C are as follows:

 
ci c1 c2 c3 c4 c5
pi 0.3 0.05 0.1 0.3 0.25

If the cells in C are partitioned into two paging zones Z1 = {c1c2c3}, Z2 = {c4c5}, the average cost of paging, , and the average paging delay in locating the mobile, , are:

 = n1 + (n1 + n2) = 3(0.3 + 0.05 + 0.1) + (3 + 2)(0.3 + 0.25) = 3 x 0.45 + 5 x 0.55 = 4.1

 = 1 +2 = 1(0.3 + 0.05 + 0.1) + 2(0.3 + 0.25) = 1 x 0.45 + 2 x 0.55 = 1.55

If the cells in C are partitioned into two paging zones Z1 = {c1c4}, Z2 = {c2c3c5}, the average cost of paging, , and the average paging delay in locating the mobile, , are:

 = n1 + (n1 + n2) = 2(0.3 + 0.3) + (3 + 2)(0.05 + 0.1 + 0.25) = 2 x 0.6 + 5 x 0.4 = 3.2

 = 1 +2 = 1(0.3 + 0.3) + 2(0.05 + 0.1 + 0.25) = 1 x 0.6 + 2 x 0.4 = 1.4

Given the number of cells in a location area C, the probabilities of each cells that a mobile is located at the cell, and the fixed number w of paging zones, write a program to partition the cells in C into w paging zones such that the average cost of paging to find the location of the mobile is minimized.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases Tis given in the first line of the input. Each test case consists of two lines. The first line of each test case contains two integers. The first integer, n, is the number of cells in a location area, and the second integer, w, is the number of paging zones, where 1wn100. The second line of each test case contains nintegers u1u2,..., un, where the probability pi for each cell ci in C is pi = ui/(u1 + u2 + ... un). All integers in the second line are between 1 and 10,000.

Output

Your program is to write to standard output. Print exactly one line for each test case. The line should contain the minimum average cost of paging to find the location of the mobile. The output should have a precision of exactly 4 digits after decimal point. You may round to the 4 digits after decimal point or round off at the 4-th digit after decimal point.

The following shows sample input and output for two test cases.

Sample Input

2
5 2
30 5 10 30 25
5 5
30 5 10 30 25

Sample Output

3.2000
2.3000
 #include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; bool cmp(double x,double y)
{
return x>y;
} const double inf=0x3f3f3f3f; int main()
{
int i,j,k;
int n,w;
int T;
double dp[][];
double a[],sum[];
double s;
scanf("%d",&T);
while(T--)
{
s=;
scanf("%d %d",&n,&w);
for(i=;i<=n;i++)
{
scanf("%lf",&a[i]);
s=s+a[i];
}
sort(a+,a+n+,cmp);
sum[]=;
for(i=;i<=n;i++)
{
sum[i]=sum[i-]+a[i];
} dp[][]=;
for(i=;i<=n;i++)
dp[i][]=inf; for(i=;i<=n;i++)
{
for(j=;j<=i && j<=w;j++)
{
dp[i][j]=inf;
for(k=;k<i;k++)
{
if(j-<=k)
dp[i][j]=min(dp[i][j],dp[k][j-]+i*(sum[i]-sum[k])/s);
}
}
} printf("%.4lf\n",dp[n][w]);
}
return ;
}

UVA 1456 六 Cellular Network的更多相关文章

  1. Educational Codeforces Round 15 C. Cellular Network(二分)

    C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard ...

  2. Educational Codeforces Round 15 Cellular Network

    Cellular Network 题意: 给n个城市,m个加油站,要让m个加油站都覆盖n个城市,求最小的加油范围r是多少. 题解: 枚举每个城市,二分查找最近的加油站,每次更新答案即可,注意二分的时候 ...

  3. Codeforces Educational Codeforces Round 15 C. Cellular Network

    C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard ...

  4. cf702C Cellular Network

    C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard ...

  5. Educational Codeforces Round 15_C. Cellular Network

    C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard ...

  6. codeforces 702C Cellular Network 2016-10-15 18:19 104人阅读 评论(0) 收藏

    C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard ...

  7. codeforces 702C C. Cellular Network(水题)

    题目链接: C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input st ...

  8. 深度学习(二十六)Network In Network学习笔记

    深度学习(二十六)Network In Network学习笔记 Network In Network学习笔记 原文地址:http://blog.csdn.net/hjimce/article/deta ...

  9. CodeForce-702C Cellular Network(查找)

    Cellular Network CodeForces - 702C 给定 n (城市数量) 和 m (灯塔数量): 给定 a1~an 城市坐标: 给定 b1~bm 灯塔坐标: 求出灯塔照亮的最小半径 ...

随机推荐

  1. 夺命雷公狗jquery---4内容选择器

    <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...

  2. ETM and PTM

    ETM:embedded Trace Macrocell PTM:Program Flow Trace Macrocell ETM-A7 macrocell提供Cortex-A7 MPcore的ins ...

  3. [php] PHPExcel插入图片

    其它的代码就不贴了,直接上关键代码: $objPHPExcel = new PHPExcel(); $objPHPExcel->setActiveSheetIndex(0); $objActSh ...

  4. SQL 中常见的系统存储过程

    -- 来源于网络 -- 更详细的介结参考联机帮助文档 xp_cmdshell --*执行DOS各种命令,结果以文本行返回. xp_fixeddrives --*查询各磁盘/分区可用空间 xp_logi ...

  5. Java高效编程之一【创建和销毁对象】

    一.考虑用静态工厂方法替代构造函数 代表实现:java.util.Collection Framework Boolean类的简单例子: public static Boolean valueOf ( ...

  6. tomcat支持的websocket服务

    首发:个人博客 在tomcat7之后的版本,写个websocket服务程序非常容易——如以下代码所示,当客户端建立了一个连接并发送了一些什么内容到服务器,服务器将每隔两秒返回一个字符串“world”. ...

  7. 影响SQL server性能的关键

    一.逻辑数据库和表的设计数据库的逻辑设计.包括表与表之间的关系是优化关系型数据库性能的核心.一个好的逻辑数据库设计可以为优化数据库和应用程序打下良好的基础.   标准化的数据库逻辑设计包括用多的.有相 ...

  8. android 项目学习随笔十五(ShareSDK开放平台)

    ShareSDK开放平台http://www.mob.com/#/

  9. android 学习随笔七(网络:图片及文本传输及线程关系 )

    主线程.子线程.UI的关系 简单的HTTP请求 -------------------------------------------------------- public class MainAc ...

  10. Error -26612: HTTP Status-Code=500 (Internal Server Error) ...

    造成HTTP-500错误,有朋友告诉我如下几个可能: 1.运行的用户数过多,对服务器造成的压力过大,服务器无法响应,则报HTTP500错误.减小用户数或者场景持续时间,问题得到解决. 2.该做关联的地 ...