UVA 1456 六 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 = {c1, c2,..., 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 Z1, Z2,..., 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 = {c1, c2,..., 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 = {c1, c2, c3}, Z2 = {c4, c5}, 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 = {c1, c4}, Z2 = {c2, c3, c5}, 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 1
w
n
100. The second line of each test case contains nintegers u1, u2,..., 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
- 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 ...
- 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 ...
- 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 ...
- cf702C Cellular Network
C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard ...
- Educational Codeforces Round 15_C. Cellular Network
C. Cellular Network time limit per test 3 seconds memory limit per test 256 megabytes input standard ...
- 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 ...
- 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 ...
- 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 ...
- 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
- Power Gating的设计(架构)
switching network的层次: 一般选择flatted的形式,hierarchy的结构对voltage drop和performance delay有影响. Power network的结 ...
- scan & ATPG
Testability用来表征一个manufactured design的quality. 将testability放在ASIC前端来做,成为DFT(Design For Test),用可控(cont ...
- JSP-07-使用JavaBean封装数据
7.1 常命包名 Dao 包中的接口(NewsDao)以及类(NewsDaoImpl)注意负责和数据操作相关的事情. Service 包中的接口和类对dao的方法进行封装和调用,注意负责和业务逻辑相关 ...
- 嵌套错误Inline markup blocks (@<p>Content</p>) cannot be nested. Only one level of inline markup is allowed
例子: @{Html.Telerik().Splitter().Name("MainSplitter") .Orientation(SplitterOrientation.Vert ...
- 为 Macbook 增加锁屏热键技巧
第一步,找到“系统偏好设置”下的“安全性与隐私”,在“通用”页里勾上“进入睡眠或开始屏幕保护程序后立即要求输入密码”. 第二步,要用快捷键启动屏幕保护程序,相对复杂一点.在“应用程序”里找到“Auto ...
- TI CC2541增加一个可读写, 又可以Notify的特征字
参考这个博客: http://blog.csdn.net/feilusia/article/details/48235691 值得注意是, 测试前, 在手机中先取消对原有的设备的配对.
- Linux系统中“动态库”和“静态库”那点事儿【转】
转自:http://blog.chinaunix.net/uid-23069658-id-3142046.html 今天我们主要来说说Linux系统下基于动态库(.so)和静态(.a)的程序那些猫腻. ...
- Mysql String Functions
SUBSTRING_INDEX(str,delim,count) 按标识符截取指定长度的字符串 mysql); -> 'www.mysql' mysql); -> 'mysql.com' ...
- PHP获取不了React Native Fecth参数的解决办法代码是怎样?
fetch('https://mywebsite.com/endpoint/', { method: 'POST',headers: {'Accept': 'application/json','Co ...
- 11、Jsp加强/EL表达式/jsp标签
1 Jsp基础回顾 Jsp基础 1)Jsp的执行过程 tomcat服务器完成:jsp文件->翻译成java文件->编译成class字节码文件-> 构造类对象-> 调用方法 to ...