hdu2955-Robberies(01背包)

最新推荐文章于 2022-10-28 22:02:50 发布

转载最新推荐文章于 2022-10-28 22:02:50 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/wangtao971115/p/10358311.html

本文探讨了一个基于概率的银行抢劫模拟问题，通过动态规划算法计算在不被捕获的概率限制下，如何最大化抢劫收益。文章提供了一段C++代码实现，详细解释了如何设置背包容量为钱数，值为抢到该值钱下逃脱的概率，并最终找到在限定概率下最大可抢金额。

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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

Sample Output

2
4
6

一般01背包是以体积为背包容量,但是这里概率是浮点数,不是很方便

呢么我们可以将背包容量设置为钱数,值为抢到该值钱下逃脱的概率

最后从大到小遍历,可以逃脱就输出

AC:

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{
    double m,p[101],dp[10001];//p为概率,dp为枪i百万钱下逃脱的概率
    int t,w[101],n,sum;//w为银行可以抢的钱数
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%lf %d",&m,&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d %lf",&w[i],&p[i]);
            sum+=w[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=0;i<n;i++)         //计算出抢j百万钱下的最大不被抓概率
            for(int j=sum;j>=p[i];j--)
                dp[j]=max(dp[j],dp[j-w[i]]*(1-p[i]));

        for(int i=sum;i>=0;i--)       //然后从大到小遍历,不被抓概率大于要求就输出
        {
            if(dp[i]>1-m)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/wangtao971115/p/10358311.html