hdu 2955 Robberies

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24144    Accepted Submission(s): 8915

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

 

给出一个被抓住的最小概率 问不被抓住的情况下 能偷多少钱


这道题应该反过来想，用dp[i]存偷了i时的没被抓的概率，这个时候钱就是容量，概率就是价值，然后用01背包，再倒序遍历(因为dp数组此时正序是非递减的)，找到第一个dp[i]是大于1-被抓最小概率的，输出i就行了。

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
struct st{
    int c;
    double p;
};
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        double dp[11111];
        double m;
        scanf("%lf%d",&m,&n);
        m=1-m;
        int i,j,k;
        int sum=0;
        st st1[11111];
        for(i=0;i<n;i++)
        {
        scanf("%d%lf",&st1[i].c,&st1[i].p);
            st1[i].p=1-st1[i].p;
            sum+=st1[i].c;
        }
        memset(dp,0,sizeof(dp));
        dp[0]=1.0;
        for(i=0;i<n;i++)
        for(j=sum;j>=st1[i].c;j--)
        dp[j]=max(dp[j],dp[j-st1[i].c]*st1[i].p);
        
        for(i=sum;i>=0;i--)
        {
            if(dp[i]-m>0)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}