HDU--dp练习--1001--Robberies

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

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

题目大意：

强盗偷东西。偷东西会被捕。每家银行都会有其价值和强盗被捕的概率。求其在不被捕的情况下，所能偷得的最大价值。

题目分析：

01背包问题。

被捕的概率P可以取其不被捕的概率进行计算。即逃跑概率为1-P。

用被捕的概率计算比较麻烦……没有尝试出来。

用不被捕的概率进行计算，dp[0] = 1，下标表示偷得的价值。其余均为0.

源代码如下：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int money[100005],i,j,N,n,sum;//money表示每家银行的价值。
    double m,po[100005],dp[100005];         //po表示被捕的概率
    cin >> N;
    while (N--)
    {
        sum = 0;
        cin >> m >> n;
        memset(money,0,sizeof(money));
        memset(po,0,sizeof(po));
        memset(dp,0,sizeof(dp));
        for (i = 1; i <= n; i++)
        {
            cin >> money[i] >> po[i];
            sum += money[i];
        }
        for (i = 1; i <= n; i++)//以不被捕的概率进行计算
            po[i] = 1 - po[i];
        m = 1 - m;
        dp[0] = 1;//不偷东西的情况下不被捕的概率为1；
        for (i = 1; i <= sum; i++)
            dp[i] = 0;
        for (i = 1;i <= n; i++)//01背包问题，求不被捕的概率。
        {
            for (j = sum; j >= money[i]; j--)
                dp[j] = max(dp[j],dp[j-money[i]] * po[i]);
        }
        //for (i = 0; i <= sum; i++)
          //  cout << dp[i] << endl;
        for (i = sum; i >= 0; i--)  //在不被捕的概率下，最多能偷到多少家银行。
        {
            if (dp[i] - m > 0.000000001)
                break;
        }
        cout << i << endl;
    }
    return 0;
}