HDU2955 Robberies

Robberies

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

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

 

题意：

一小偷要偷银行，想要被抓的几率不大于p的前提下偷到的价值最大，给出每家银行的价值和被抓的概率，求最优解。

题解：

因为给出的几率为浮点型且精度比较高，所以不能将浮点数转化为整型做，所以dp要用来保存安全概率，p为被抓概率，那么1-p就是安全的概率。状态方程：

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

AC代码：

#include
using namespace std;
int t,n,m,v[105],sum;
double w[105],p,dp[10005];
int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--){
        cin>>p>>n;p=1-p;sum=0;
        for(int i=1;i<=n;i++)cin>>v[i]>>w[i],sum+=v[i];
        memset(dp,0,sizeof dp);dp[0]=1;
        for(int i=1;i<=n;i++)
            for(int j=sum;j>=v[i];j--)
            dp[j]=max(dp[j],dp[j-v[i]]*(1-w[i]));
        for(int i=sum;i>=0;i--)
            if(dp[i]-p>0.000000001){cout<