zoukankan      html  css  js  c++  java
  • HDOJ 2955 Robberies

    类似01背包的DP

    Robberies

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 7013    Accepted Submission(s): 2630


    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.

    HDOJ 2955 Robberies - qhn999 - 码代码的猿猿

    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 <iostream>
    #include <cstring>

    using namespace std;

    double f[100000];

    int main()
    {
        int T;
        cin>>T;
    while(T--)
    {
        int sum=0;
        int m[101];
        double sur[101];
        memset(f,0,sizeof(f));
        f[0]=1;
        int N; double P;
        cin>>P>>N;
        for(int i=0;i<N;i++)
        {
            double k;
            cin>>m>>k;
            sur=1-k;
            sum+=m;
        }

        for(int i=0;i<N;i++)
        {
            for(int j=sum;j>=m;j--)
                f[j]=max(f[j],f[j-m]*sur);
        }

        P=1-P;
        for(int j=sum;j>=0;j--)
            if(f[j]>=P)
            {
                cout<<j<<endl;
                break;
            }
    }
        return 0;
    }



  • 相关阅读:
    构造代码块重要理解
    Java中静态代码块、构造代码块、构造函数、普通代码块
    MySQL-分组查询(GROUP BY)及二次筛选(HAVING)
    mysql select将多个字段横向合拼到一个字段
    java语言支持的变量类型
    static修饰属性,方法,类
    恶意代码分析----网络环境配置
    Windows反调试技术(下)
    Windows反调试技术(上)
    脱壳入门----常见的寻找OEP的方法
  • 原文地址:https://www.cnblogs.com/CKboss/p/3351061.html
Copyright © 2011-2022 走看看