01背包 HDU - 2955 小偷抢银行

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#01背包 HDU - 2955 小偷抢

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本文介绍了一个基于01背包问题的算法案例，通过计算不同银行抢劫的风险与收益比，找到最大收益的同时确保被抓风险低于可接受水平的最佳方案。

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 <iostream>
#include  <algorithm>
#include <cstring>
using namespace std;
int t,n;
double p;
double dp[10005];
int money[105];
double pro[105];
int main()
{
	cin>>t;
	while(t--)
	{
		int sum=0;
		cin>>p>>n;
		for(int i=1;i<=n;i++)
		{
			double d;
			cin>> money[i]>>pro[i];
			sum+= money[i];
		}
		 memset(dp,0,sizeof(dp));
		 dp[0]=1;  //dp里面为不被抓的概率  这样表示什么都不拿 那么不被抓的概率一定为1

		for(int i=1;i<=n;i++)
		for(int j=sum;j>= money[i];j--)
		{
			dp[j]=max(dp[j],dp[j- money[i]]*(1-pro[i])); //找最大的不被抓的概率
		}

		for(int i=sum;i>=0;i--)
		{
			if(dp[i]>=(1-p))  {
				cout<<i<<endl;
				break;
			}
		}

	}
}


/*题意：
给N个银行，给出每个银行里相应的钱数money[i]和抢这个银行被抓的概率pro[i]。
问在不被抓的情况下(被抓的概率小于给定值p时安全)能抢得的最大钱数。*/

/*思路：
1、抢n个银行，不被抓，是每次都不被抓，用乘法原理处理每次不被抓的概率。
2、转化为0 1 背包：将所有银行的总钱数当做背包的容积v，将不被抓的概率当做价值，
       在一般的0 1背包中都是加上价值 ，这里由于是乘法原理，所以应该是转化为乘法。
       这样，传统的01背包累加问题变成了垒乘。
3、状态转移方程：dp[j]=max(dp[j],dp[j- money[i]]*(1-pro[i]))
     dp[j]表示抢j钱时的最大不被抓概率。*/