hdu 2955 Robberies

Myth丶恋晨 2022-06-16 01:11 116阅读 0赞

# Robberies #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 23305 Accepted Submission(s): 8592**

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
      
       
      
       
    
      
       
      
       
       
        题目大意：
       
        抢银行，对于每一个给定的金额都会有一个被容忍的可能性，只有总的可能性低于被发现的可能性，才不会被抓到，他妈妈认为如果他们一起抢劫的银行的概率低于这个水平，他就足够安全了。（嘻嘻，抢银行被发现的可能很大哦）。
      
       
      
       
       
        	第一行输入一个T，代表测试组数，对于没个情景的第一行输入一个浮点数P，代表被发现的概率，和一个N代表他计划的银行数目，接下来的N行，每行两个数，包含抢的金额Mj和被发现的概率Pj，输出可以抢到的最大金额数，越多越好哟。
      
       
      
       
       
        
    
      
       
      
       
       
        题目解析：这个题首先发现的是安全的概率是一定也就相当于背包的容量是一定的，那么就要输出最大的金额，显然是一个01背包问题。
      
       
      
       
       
        	  由题意可以写出转移方程：
       
        dp[j]=max(dp[j],dp[j-stru[i].money]*(1-stru[i].chance)); 
      
       
      
       
       
        
    
      
       
      
       
        
       代码：
      
       
      
       
       
        #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    double dp[10010];
    struct node
    {
        int money;
        double chance;
    }stru[10010];
    int main()
    {
        int T,i,j,n,sum;
        double P;
        cin>>T;
        while(T--)
        {
            cin>>P>>n;
            sum=0;
            for(i=0;i<n;i++)
            {
                cin>>stru[i].money>>stru[i].chance;
                sum+=stru[i].money;
            }
            memset(dp,0,sizeof(dp));
            dp[0]=1;
            for(i=0;i<n;i++)
                for(j=sum;j>=stru[i].money;j--)
                dp[j]=max(dp[j],dp[j-stru[i].money]*(1-stru[i].chance));
            for(i=sum;i>=0;i--)
            if(dp[i]>1-P)
                    {
                        cout<<i<<endl; break;
                    }
    
    
        }
        return 0;
    }

[con211-1010-1.jpg]: /images/20220616/e512a84bbbcb40b5bc6e48d0bc19c535.png