HDU 2955（01背包问题）

M - 01背包

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

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

 

 

题解： 

        一个人想要抢劫银行，每家银行多有一定的金额和被抓到的概率，知道Roy被抓的最大概率P，求他在不被被抓的情况下，抢劫的钱最多。

        被抓概率为pj,安全概率1-pj，他的安全概率大于1-P时都是安全的。抢劫的金额为0时，肯定是安全的，所以dp[0]=1;其他金额初始为最危险的所以概率全为0；

思路：

        和背包问题做一个简单的转化：把每个银行的储钱量之和total当成背包容量C，然后不被抓的概率1-pj当成价值w来求。那么他的最大的不被抓的概率下抢劫的钱最多的情况就是乘积了，最后再枚举一下就行了，为什么要用乘积呢这里，想一想数学中涉及到概率的问题，而不是简单概率的相加吧，。这里计算的是抢劫到j钱不被抓的概率，相乘才能保存下来。

dp[ j] 表示在抢劫到钱j时不被抓概率。最后，只要找出满足要求的最大j。

代码及案例显示

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int T,n,total;
double p;
double a[],dp[];
int b[];
int main()
{
    cin>>T;
    while(T--)
    {
        total=;
        cin>>p>>n;
        for(int i=; i<n; i++)
        {
            cin>>b[i]>>a[i];
            total+=b[i];
        }
        memset(dp,,sizeof(dp));
        dp[]=;                        //没抢到钱时，逃脱的概率为1
        for(int i=; i<n; i++)
            for(int j=total; j>=a[i]; j--)
            {
                dp[j]=max(dp[j],dp[j-b[i]]*(-a[i]));
                //cout<<d[j]<<" "<<j<<endl;
            }
        for(int i=total; i>=; i--)
        {
            if(dp[i]>-p)
            {
                cout<<i<<endl;
                break;
            }
        }
    }
    return ;
}

下面是第一组案例的各种抢劫钱数的概率