hdu2955 dp（背包）

最新推荐文章于 2021-04-14 23:07:27 发布

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本文介绍了一个基于概率的算法，用于解决一个特殊的“银行抢劫”问题。该算法通过动态规划求解在不超过被抓概率限制的情况下，能获取的最大金额。具体探讨了如何计算多个独立事件不发生（即抢劫不被发现）的概率乘积，并给出了实现这一目标的详细代码。

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Robberies

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

思路：
这题的样例出的非常坑，一开始让人以为只要概率相加小于P即可。于是将概率放大1000倍，用01背包来做。直接wa了。
但是被抓的概率不是简单的直接相加。于是考虑不被抓的概率，假设被抓概率为p1，p2，于是不被抓的概率为（1-p1）*（1-p2）。
我们只要求偷掉价值w时不被抓的最大概率。
用dp来做。
dp[i] 表示偷掉w百万元时不被抓的最大概率。
状态转移方程：dp[j]=max(dp[j],dp[j-w[i]*(1-p[i]));

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=10005;
double dp[N];
int w[105];double p[105];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;double P;
        scanf("%lf%d",&P,&n);
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%lf",&w[i],&p[i]);
            sum+=w[i];
        }
        for(int i=0;i<=sum;i++)dp[i]=-1;
        dp[0]=1;//不偷钱肯定不被抓
        for(int i=1;i<=n;i++)
        {
            for(int j=sum;j>=w[i];j--){
                dp[j]=max(dp[j],dp[j-w[i]]*(1.0-p[i]));
            }
        }
        for(int j=sum;j>=0;j--){
            if(dp[j]>0&&1-dp[j]<=P)
                {
                printf("%d\n",j);break;
                }
        }
    }
    return 0;
}