【基础dp】Robberies

A - Robberies

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 
         
         
         
         

开一个一位dp数组 下标表示钱数  ，里面存放抢了这么多钱时最大的逃脱概率。

状态转移方程为dp[j]=max(dp[j],dp[j-c[i]]*(1-b[i]));

最后查找一下 满足1-dp[i]<p的最大的i

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
int a[10700];
double dp[10700];
double b[10700];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double p;
        int n;
        scanf("%lf%d",&p,&n);

        int mj;
        double pj;
        int sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%lf",&a[i],&b[i]);
           sum+=a[i];
        }

        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=0;i<n;i++)
        {
            for(int j=sum;j>=a[i];j--)
                dp[j]=max(dp[j],dp[j-a[i]]*(1-b[i]));
        }

       int ans=0;
        for(int i=sum;i>=0;i--)
        {
            if(1-dp[i]<=p)
            {
                ans=i;
                break;
            }
            //printf("%d\n",dp[sum]);
        }
        printf("%d\n",ans);
    }
    return 0;
}