Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32227    Accepted Submission(s): 11698

 

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

問題連結：HDU2955 Robberies

問題描述：搶銀行，有N個銀行，每個銀行有兩個值，擁有的錢數和搶劫此銀行被抓的概率。在被抓概率小於P時，求能搶到的最多錢數。注意被抓概率不是簡單求和，而是P=1-（1-p1）*(1-p2)*（1-p3），p1,p2,p3是搶劫各個銀行被抓的概率

解題思路：01揹包問題。使用各個銀行的總錢數作為揹包容量，概率為價值，dp[v]儲存搶劫錢數v的最大安全概率。問題轉化為安全概率大於等於1-P時的最大容量

AC的C++程式碼：

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int N=105;

struct Bank{
	int v;//錢數 
	double p;//被抓概率 
}a[N];

double dp[10100];

int main()
{
	int t,n;
	double P;
	scanf("%d",&t);
	while(t--){
		memset(dp,0,sizeof(dp));
		int sum=0;
		scanf("%lf%d",&P,&n);
		for(int i=0;i<n;i++){
			scanf("%d%lf",&a[i].v,&a[i].p);
			sum+=a[i].v;
		}
		dp[0]=1;//不偷的安全概率為1 
		for(int i=0;i<n;i++)
		  for(int j=sum;j>=a[i].v;j--)
		    dp[j]=max(dp[j],dp[j-a[i].v]*(1-a[i].p));
		for(int i=sum;i>=0;i--)
		  if(dp[i]>=1-P){
		  	printf("%d\n",i);
		  	break;
		  }
	}
	return 0;
 }