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分手后的思念是犯贱 2022-08-03 13:41 71阅读 0赞

# Robberies #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 16258    Accepted Submission(s): 5954**

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.  
  
![con211-1010-1.jpg][]  
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

Source

[IDI Open 2009][]  
  
  
  
  
  
01背包，把总的钱数看作背包容量，把每个银行的钱数看作物品重量，把每个银行的不被抓到风险看作物品的价值，其中抢劫多家银行的时候要算不被抓的概率相乘，而不是被抓的概率相加（有两家银行，抢每一家的概率都是0.5，两家都抢就一定被抓？）。

#include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;
        while((c=getchar())<'0' || c>'9');
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res;
    }
    struct st
    {
        int val;
        double p;
    }a[111];
    double dp[111111];
    
    
    int main()
    {
        int T=in();
        while(T--)
        {
            double p;
            scanf("%lf",&p);
            int n=in();
            int sum=0;
            for(int i=1;i<=n;i++)
            {
                a[i].val=in();
                sum+=a[i].val;
                scanf("%lf",&a[i].p);
            }
            mem(dp,0);
            dp[0]=1;
    
            for(int i=1;i<=n;i++)
            {
                for(int j=sum;j>=a[i].val;j--)
                {
                    dp[j]=max(dp[j],dp[j-a[i].val]*(1-a[i].p));
                }
            }
            for(int i=sum;i>=0;i--)
            {
                if(dp[i]>1-p)
                {
                    printf("%d\n",i);
                    break;
                }
            }
    
        }
    }

[con211-1010-1.jpg]: /images/20220731/673b6fcd38e7438287e2e5fd0be3e073.png
[IDI Open 2009]: http://acm.hdu.edu.cn/search.php?field=problem&key=IDI+Open+2009&source=1&searchmode=source