Robberies（HDOJ2955)

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7620    Accepted Submission(s): 2859

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
这个题目是一个01背包的问题，但是使用正常的思路会发现，数组名会会粗线小数，这显然不科学，所以，应该换种思路，讲价值作为数组名，讲套牌概率存到数组当中，同时，因为存储的是概率，要注意的是概率的计算式乘除，在初始化dp数组时，应该注意将dp[0]初始化为1，因为没拿到任何价值的话，逃跑率必为1
代码：
#include<iostream>using namespace std;double dp[10005],p[105];int m[105];int main(){int T,n,i,j,sum;double z;cin>>T;while(T--){cin>>z>>n;z=1.0-z;sum=0;for(int i=1;i<=n;i++){cin>>m[i]>>p[i];p[i]=1-p[i];sum=sum+m[i];}for(i=0;i<=sum;i++){dp[i]=0.0;}dp[0]=1.0;for(i=1;i<=n;i++){for(j=sum;j>=m[i];j--){dp[j]=dp[j-m[i]]*p[i] > dp[j] ? dp[j-m[i]]*p[i]:dp[j];}}for(j=sum;j>=0;j--){if(dp[j]-z>0.000000001){cout<<j<<endl;break;}}}return 0;}