动态规划：HDU-2955-0-1背包问题：Robberies

解题心得：

1. 这题涉及概率问题，所以要运用概率的知识进行解答。题目要求不被抓到的概率，但是给出的是被抓到的概率，所要用1减去后得到答案。最好使用double类型，避免精度问题导致WA。
2. 先算出可以抢劫的总钱数，以此动态规划。

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Robberies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 23012 Accepted Submission(s): 8489

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

#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
const int maxsize = 10010;
struct ai
{
    double p;
    int v;
}a[maxsize];
int main()
{
    long long totle;
    double d[maxsize];
    int t;
    cin>>t;
    while(t--)
    {
        totle = 0;

        int n;
        double pro;
        cin>>pro>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i].v>>a[i].p;
            totle += a[i].v;
        }
        d[0] = 1;//这个初始化很重要。
        for(int i=1;i<=totle;i++)
            d[i] = 0;
        for(int i=1;i<=n;i++)
        {
            for(int j=totle;j>=a[i].v;j--)
            {
                d[j] = max(d[j],d[j-a[i].v]*(1-a[i].p));
            }
        }
        for(int i=totle;i>=0;i--)
        {
            if(d[i] >= (1-pro))
            {
                cout<<i<<endl;
                break;
            }
        }
    }
    return 0;
}