HDU Robberies 2955 dp

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23310    Accepted Submission(s): 8595

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

  

   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

 

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题意：盗贼想去抢劫几家银行，但是不想被抓，每抢劫一家银行有不同的被抓的可能，只要不超过某个极限值就不会被抓

实例分析

第一行的3代表三个实例，然后是

（极限被抓概率） （银行数量）

（银行有多少钱） （这次被抓概率）

大体就是这样，做题的时候有两点需要注意

（1）怎么建立0-1背包模型，第一层for循环一定是循环的物品，但是第二层循环怎么搞，浮点数没给出来有多少位小数，所以不能进行循环，那么只有银行的钱数了，我们以银行的钱数进行循环，然后对应的值为概率

（2）dp里面对应的值为概率，但是不能为被抓的概率，得是不被，抓的概率，每一次都是不被抓的概率相乘，为什么呢？上便了度娘也没找到个我能看懂的解释（或许是我太菜了），我觉得可以这么想“每多强一个银行，在原来不被抓的基础上就得*接下来不被抓的概率”，如果只是进行被抓的概率相加，那么都有可能超过一，想想是不是这个道理。当然，这只是本菜杜撰的，如有大神经过，欢迎留言。

ac代码:

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

using namespace std;

int n,m,val[105],sum;
double mom,fx[105];
double dp[100005];

int main(){
	
	scanf("%d",&n);
	while(n--){
		sum=0;
		scanf("%lf%d",&mom,&m);
		for(int i=1;i<=m;i++){
			scanf("%d%lf",&val[i],&fx[i]);
			sum+=val[i];
		}
		for(int i=0;i<=sum+1;i++){
			dp[i]=0.0;
		}
		dp[0]=1;
		for(int i=1;i<=m;i++){
			for(int j=sum;j>=val[i];j--){
				dp[j]=max(dp[j],dp[j-val[i]]*(1-fx[i]));
			}
		}
		
		for(int i=sum;i>=0;i--){
			if((1-dp[i]) <= mom)
			{
				printf("%d\n",i);
				
				break ;
			}
		}
	}
	
	
	return 0;
}