HDU-2955 Robberies

           01背包的概率问题

        http://acm.hdu.edu.cn/showproblem.php?pid=2955    

  状态方程 ：dp[j]=max(dp[j],dp[j-m[i]]*1-pi[i]);

  思路是：  

当前的概率基于前一种状态的概率，即偷n家银行而不被抓的概率等于偷n-1家银行不被转的概率乘以偷第n家银行不被抓的概率。

用dp[i]表示偷价值为 i 时不被抓的概率，则状态转移方程为：

          dp[j] = max(dp[j] , dp[j-m[i]] * (1-p[i]));

                              Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9334    Accepted Submission(s): 3489

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

#include <iostream> 
#include <algorithm>
using namespace std;
int main()
{
	int i,t,n,sum,j;
	double p;
	int  m[200];
   double pi[200],dp[10001];
     cin>>t;
	 while(t--)
	 {
		  sum=0;
		 cin>>p>>n;
		 p=1-p;
         for(i=0;i<n;i++)
		 { 
		    cin>>m[i]>>pi[i];
             sum+=m[i];
			 pi[i]=1-pi[i];
		 }
		 memset(dp,0,sizeof(dp));
		 dp[0]=1;
		 for(i = 0;i < n;i++) 
            for(j = sum;j >= m[i];j--)
                if(dp[j]<dp[j-m[i]]*pi[i])
				  dp[j] = dp[j-m[i]]*pi[i];
		    for(i = sum;i >= 0;i--)
		      if(dp[i]>p)
			  {
				  cout << i << endl;
			      break;
			  }
			  
     }
	 return 0;
}