01背包M

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

 

 

思路：

     把被抓的概率转化为安全逃脱的概率p=1-p

     再用循环从安全概率里面找到题目要求的概率范围

 

     状态转移方程：d[j]=max(d[j],d[j-cost[i]]*gl[i])

     d[j]表示小偷偷走j钱逃脱的概率

    

     总的步骤就是：输入，转化被抓的概率为安全概率（方便些），找每种情况的概率，找最终的临界值

 

 

源代码：

    

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 10000+10   //数组大小要注意，选择最大的情况，100家银行
int cost[maxn];
double gl[maxn],d[maxn];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,sum=0;
        double p;
        cin>>p>>n;
        p=1-p;            //逃脱概率
        for(int i=0;i<n;i++)
        {
            cin>>cost[i]>>gl[i];
            gl[i]=1-gl[i];      //抢劫每家银行的逃脱概率     
            sum+=cost[i];       //抢劫所有的银行的金额
        }
        for(int i=0;i<=sum;i++)
               d[i]=0.0;
                d[0]=1.0;           //初始化小偷不偷东西，为最安全
        for(int i=0;i<n;i++)        //不等于n，等于n就错了
        {
            for(int j=sum;j>=cost[i];j--)
            {
                d[j]=max(d[j],d[j-cost[i]]*gl[i]);  //小偷偷j钱逃脱的最大概率
            }
        }
        for(int i=sum;i>=0;i--)
        {
            if(d[i]>=p) //可以等于p   找到临界值
            {
                cout<<i<<endl;
                break;
            }
        }
    }

 
    return 0;
}

 

心得：

  比之前的01背包N难点，之前就是加法运算，但是这个概率要相乘，看题目的案例还以为又是加来加去，这个的话后面还要找临界值，好好理解下吧！！！