hdu2955（01背包）

HDU_2955 Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 20096    Accepted Submission(s): 7443

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

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

 

Sample Output

246

其实挺像裸01背包，但是如果把概率当做容量，金额当做价值，可以发现下标是浮点数

虽然可以乘以一个10的多次幂，但是该数不好求，而且也容易内存过大。 

于是

可以假设可以偷得总金额为容量，价值表示没有被抓到的概率。

dp公式可以表示为：

dp[j]=max(dp[j],dp[i-v[i]]*(1-w[i]));

v[i]表示金额

w[i]表示概率

#include <iostream>using namespace std;const int maxn=105;int m[maxn];double p[maxn],dp[maxn*maxn];int main(){//    freopen("data.in.txt","r",stdin);    int T;    cin>>T;    while(T--){        double P;        int N;        cin>>P>>N;        int sum=0;        for(int i=0;i<N;i++){            cin>>m[i]>>p[i];            sum+=m[i];            //            cout<<p[i]<<endl;        }                dp[0]=1;        for(int i=1;i<=sum;i++)            dp[i]=0;        for(int i=0;i<N;i++)            for(int j=sum;j>=m[i];j--)                dp[j]=max(dp[j],dp[j-m[i]]*(1-p[i]));                        for(int i=sum;i>=0;i--){//            cout<<dp[i]<<endl;            if(dp[i]>1-P){                cout<<i<<endl;                break;            }        }            }     return 0;    }