10397780 2014-03-26 00:13:51 Accepted 2955 46MS 480K 676 B C++ 泽泽

http://acm.hdu.edu.cn/showproblem.php?pid=2955

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9836    Accepted Submission(s): 3673

Problem Description
The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.


For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.

His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.

 
Input
The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj . 
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
 
Output
For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.

Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

 
Sample Input
3
0.04 3
1 0.02
2 0.03
3 0.05
0.06 3
2 0.03
2 0.03
3 0.05
0.10 3
1 0.03
2 0.02
3 0.05
 
Sample Output
2
4
6
 
 #include<stdio.h>
#include<string.h>
int main()
{
int v[];
double f[],w[];
int i,j,n,m;
int T;
scanf("%d",&T);
while(T--)
{
double rate;
int sum=;
scanf("%lf %d",&rate,&n);//rate被抓的概率,1-rate逃脱的概率
for(i=;i<n;i++)
{
scanf("%d %lf",&v[i],&w[i]);
sum+=v[i];
}
memset(f,,sizeof(f));
f[]=;//没偷逃脱的概率为1
for(i=;i<n;i++)
{
for(j=sum;j>=v[i];j--)
{
f[j]=f[j]>f[j-v[i]]*(-w[i])?f[j]:f[j-v[i]]*(-w[i]);//选择最大逃脱概率
}
}
for(i=sum;i>=;i--)
{
if(f[i]>=-rate)
{
printf("%d\n",i);
break;
}
}
}
return ;
}

HDOJ 2955 Robberies (01背包)的更多相关文章

  1. HDOJ.2955 Robberies (01背包+概率问题)

    Robberies 算法学习-–动态规划初探 题意分析 有一个小偷去抢劫银行,给出来银行的个数n,和一个概率p为能够逃跑的临界概率,接下来有n行分别是这个银行所有拥有的钱数mi和抢劫后被抓的概率pi, ...

  2. hdu 2955 Robberies 0-1背包/概率初始化

    /*Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total S ...

  3. hdu 2955 Robberies (01背包)

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=2955 思路:一开始看急了,以为概率是直接相加的,wa了无数发,这道题目给的是被抓的概率,我们应该先求出总的 ...

  4. HDU 2955 Robberies(01背包变形)

    Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  5. hdu 2955 Robberies (01背包好题)

    Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  6. HDU——2955 Robberies (0-1背包)

    题意:有N个银行,每抢一个银行,可以获得\(v_i\)的前,但是会有\(p_i\)的概率被抓.现在要把被抓概率控制在\(P\)之下,求最多能抢到多少钱. 分析:0-1背包的变形,把重量变成了概率,因为 ...

  7. HDU 2955 Robberies --01背包变形

    这题有些巧妙,看了别人的题解才知道做的. 因为按常规思路的话,背包容量为浮点数,,不好存储,且不能直接相加,所以换一种思路,将背包容量与价值互换,即令各银行总值为背包容量,逃跑概率(1-P)为价值,即 ...

  8. HDU 2955 Robberies(01背包)

    Robberies Problem Description The aspiring Roy the Robber has seen a lot of American movies, and kno ...

  9. hdoj 2955 Robberies

    Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

随机推荐

  1. Linux下搭建nginx php环境

    下载安装所需包 openssl-1.0.1i.tar.gz zlib-1.2.8.tar.gz pcre-8.35.tar.gz nginx-1.7.4.tar.gz 以上为nginx依赖文件 lib ...

  2. python 中颜色的表示

    字背景颜色范围:40----49 40:黑 41:深红 42:绿 43:黄色 44:蓝色 45:紫色 46:深绿 47:白色 字颜色:30-----------39 30:黑 31:红 32:绿 33 ...

  3. iOS边练边学--cocoaPods管理第三方框架--命令行方式实现

    更换源 Gem是一个管理Ruby库和程序的标准包,它通过Ruby Gem(如 http://rubygems.org/)源来查找.安装.升级和写在软件包 gem sources --remove ht ...

  4. Java基础-设计模式之-代理模式Proxy

    代理模式是对象的结构模式.代理模式给某一个对象提供一个代理对象,并由代理对象控制对原对象的引用. 代理模式是常用的Java 设计模式,它的特征是代理类与委托类有同样的接口,代理类主要负责为委托类预处理 ...

  5. Struts2(二)---将页面表单中的数据提交给Action

    问题:在struts2框架下,如何将表单数据传递给业务控制器Action. struts2中,表单想Action传递参数的方式有两种,并且这两种传参方式都是struts2默认实现的,他们分别是基本属性 ...

  6. Spring AOP详解 、 JDK动态代理、CGLib动态代理

    AOP是Aspect Oriented Programing的简称,面向切面编程.AOP适合于那些具有横切逻辑的应用:如性能监测,访问控制,事务管理以及日志记录.AOP将这些分散在各个业务逻辑中的代码 ...

  7. ACM算法总结及刷题参考

    参考:http://bbs.byr.cn/#!article/ACM_ICPC/11777 OJ上的一些水题(可用来练手和增加自信)(poj3299,poj2159,poj2739,poj1083,p ...

  8. int方法

    代码 #int内部功能 name='Kamil.Liu' age=18 num=-11 print(dir(age)) print(age.bit_length())#返回表示当前数字占用的最少位数 ...

  9. BZOJ3626 LCA

    Description 给出一个n个节点的有根树(编号为0到n-1,根节点为0).一个点的深度定义为这个节点到根的距离+1. 设dep[i]表示点i的深度,LCA(i,j)表示i与j的最近公共祖先. ...

  10. ECSHOP Inject PHPCode Into \library\myship.php Via \admin\template.php && \includes\cls_template.php Vul Tag_PHP_Code Execute Getshell

    目录 . 漏洞描述 . 漏洞触发条件 . 漏洞影响范围 . 漏洞代码分析 . 防御方法 . 攻防思考 1. 漏洞描述 PHP语言作为开源社区的一员,提供了各种模板引擎,如FastTemplate,Sm ...