hdu2955 教你怎样抢银行划算！

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2447    Accepted Submission(s): 933

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

 

题目大意是有n家银行，每家银行可抢mi的金钱，被抓的分险是pi，当你的分险和大于p时就会被抓，求你在不被抓的前提下怎样才能抢到最多的钱

输入数据第一行是p和n，接着n行，每行两个数，分别代表mi和pi

例如，对于数据一，可以抢第二家银行，故最多可以得到2金币

这题，就是01背包的小数形式，我们可以用pi表示抢到i金币时的最大逃脱率

则就可以得到状态转移方程（详见代码）

参考代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
double q[105];
int m[105];
double dp[10005];
int main()
{
    int t ;
    cin >> t ;
    while ( t -- )
    {
        double p ;
        int n ;
        cin >> p >> n ;
        int i , j ;
        int sum = 0 ;
        for ( i = 0 ; i < n ; i ++ )
            cin >> m[i] >> q[i] , sum += m[i] ;
        memset ( dp , 0 , sizeof ( dp ) ) ;
        dp[0] = 1.0 ;
        for ( i = 0 ; i < n ; i ++ )
        {
            for ( j = sum ; j >= m[i] ; j -- )
            {
                dp[j] = max ( dp [j] , dp [j-m[i]] * ( 1 - q[i] ) ) ;
            }
        }
        for ( j = sum ;  j >= 0 ; j -- )
            if ( dp [j] >= 1 - p )
            {
                cout << j << endl;
                break;
            }
    }
    return 0;
}