hdoj 2955Robberies<背包>_hdoj-2955-优快云博客

本文链接：https://blog.youkuaiyun.com/leibniz_zhang/article/details/51099986

本文介绍了一个经典的银行抢劫问题，通过计算不同银行的抢劫收益与被抓风险，采用动态规划算法求解最大期望收益的同时确保被抓概率低于设定阈值。文章详细解释了如何将问题转化为0-1背包问题，并给出了具体的实现代码。

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Robberies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18676 Accepted Submission(s): 6890

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

Source

IDI Open 2009

开始以为概率是相加的且>0.01的,wrong的好惨，，，

用成功逃走的概率当做价值！银行的总钱数当做背包容量！

然后钱数从大到小求能逃出概率的值。。。

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
#define MA 105
int n,t,ss;
int pp[MA],qian[MA];
double bao[MA*MA],p[MA],s;
int main()
{
    scanf("%d",&t);
	while (t--)
	{
		scanf("%lf%d",&s,&n);
	    ss=0;
	    s=1.0-s;
		for (int i=0;i<n;i++)
		{
			scanf("%d%lf",&qian[i],&p[i]);
			p[i]=1.0-p[i];
			ss+=qian[i];
		}
		memset(bao,0,sizeof(bao));
		bao[0]=1;
		for (int i=0;i<n;i++)
		{
			for (int j=ss;j>=0;j--)
			{
				if (qian[i]<=j)
			    {
				bao[j]=max(bao[j],bao[j-qian[i]]*p[i]);//概率是相乘的。。。。 
				}
			}
		}
		for (int i=ss;i>=0;i--)
		{
			if (bao[i]>=s)
			{
				printf("%d\n",i);
				break;
			}
		}
    }
	return 0;
}