hdu2955的01背包问题

hdu2955

Robberies

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

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

Sample Output

246

大概意思是一只小偷去偷银行，每个银行都有一个被抓概率和一定的钱财数，求在不超过最大的被抓概率的情况下能偷到的最大钱财。

我的思路开始的时候非常鸡肋，把概率乘了一定倍数当了背包容量，宛如一个智障！！

后来才明白应该把钱财总和当作背包容量，然后在包里装的应该是概率，注意！！这个概率是当偷到 此钱财数时 所能逃跑的概率！！

 dp[j]=max(dp[j-wi[i]]*(1-pi[i]),dp[j])

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
using namespace std;

double pi[200];
int wi[200];
double dp[10005];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double p;
        int n,sum=0;
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        scanf("%lf%d",&p,&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%lf",&wi[i],&pi[i]);
            sum+=wi[i];
        }
        for(int i=0;i<n;i++)
            for(int j=sum;j>0;j--)
            {
                if(j-wi[i]>=0)
                    dp[j]=max(dp[j-wi[i]]*(1-pi[i]),dp[j]);
            }
        for(int i=sum;i>=0;i--)
        {
            if(1-dp[i]<=p)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}