HDU 2955 Robberies

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4852    Accepted Submission(s): 1844

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：Roy要偷取银行，每家银行都有现金的金额和被抓的概率，知道Roy被抓的最大概率，求在最大概率内可以偷盗的最大金额。

被抓概率可以转化为安全概率，概率的类型要为double类型。

 

状态转移方程：dp[j] = max(dp[j],dp[j-num[i]]*weigh[i])　

#include <iostream>
#include <string.h>
using namespace std;
int num[111];double weigh[111];    //num 每个银行拥有的现金  weigh 每个银行不被抓的概率
double dp[11111];              //dp[j] 偷取金额为j时逃脱被捕的概率

int main()
{
    int t;
    int n;
    int sum;
    float p;
    cin>>t;
    while(t--){
        cin>>p>>n;
        p = 1-p;
        sum=0;
        for(int i=0;i<n;i++){
            cin>>num[i]>>weigh[i];
            weigh[i] = 1 - weigh[i];
            sum+=num[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0] = 1;
        for(int i=0;i<n;i++)
            for(int j=sum;j>=num[i];j--){
                dp[j] = max(dp[j],dp[j-num[i]]*weigh[i]);
            }
        for(int i=sum;i>=0;i--){
            if(dp[i]-p>0){
                cout<<i<<endl;
                break;
            }
        }
    }
    return 0;
}