HDU 2955 Robberies

最新推荐文章于 2025-09-09 08:37:13 发布

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最新推荐文章于 2025-09-09 08:37:13 发布

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文章标签： java

本文探讨了一位名叫罗伊的虚拟银行抢劫犯如何在风险与收益之间做出明智选择，通过评估各银行的安全性和储钱量，采用贪心算法策略最大化其非法所得，同时确保被抓捕的概率低于设定阈值。利用独立概率计算，实现最优抢劫方案。

Robberies

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 6 Accepted Submission(s) : 2

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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

Sample Output

2
4
6

Source

IDI Open 2009

在这里做一个简单的转化，把每个银行的储钱量之和当成背包容量，然后

概率当成价值来求。这里是被抓的概率，我们把他转化成不被抓的概率，

然后这里的和就可以转化成乘积了，最后再枚举一下就行了。

 1 #include<stdio.h>
 2 #include<string.h>
 3 int money[101] , nkind  , sum ;
 4 float pro[101] , npro , fine[11100];
 5 int main ()
 6 {
 7     int cas , i , j , k ;
 8     scanf ( "%d" , &cas ) ;
 9     while ( cas -- )
10     {
11           sum = 0 ;
12           scanf ( "%f%d" , &npro , &nkind ) ;
13           for ( i = 0 ; i < nkind ; i ++ )
14           {
15               scanf ( "%d%f"  , money+i, pro+i ) ;
16               sum += money[i] ;
17           }
18           memset( fine , 0 , sizeof (fine) ) ;
19           fine[0] = 1 ; 
20           float  p = 1 - npro ;
21           for ( i = 0 ; i < nkind ; i ++ )
22           { 
23               for ( j = sum ; j >= money[i] ; j -- )
24                   if ( fine[j] < fine[j-money[i]]*(1-pro[i]) )
25                   {
26                        fine[j] = fine[j-money[i]]*(1-pro[i]) ;
27                   }
28           }
29           for ( i = sum ; i >= 0 ; i -- )
30               if ( fine[i] >=  p )
31               {
32                    printf ( "%d\n" , i ) ;
33                    break ;
34               }
35     }
36     return 0 ;
37 }