HDU_2955 动态规划

                                                          Robberies     

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

题意：一个强盗抢银行，每个银行都有一个 int 型存款和一个 float 型被抓的可能性，这个强盗想在被抓可能性不超过某个浮点型 K 的情况下尽可能多的抢钱，问最多能抢到多少钱？

思路：这个题在我以前做背包问题的时候做过，但是当时是将所有的浮点数放大100000倍，而且将被抓的概率直接相加，当成01背包来做。。。显而易见，WA的很惨。。。昨天开始刷动态规划46题，第一题就是这个题，问了一下暾哥，严重怀疑自己的智商。。。言归正传，我们这里将dp[ i ]作为抢到 i 元钱不被抓的概率， 然后将 n 个银行依次放入，假设第 i 个银行有钱 money[ i ] 元， 不被抓的可能性为 pos[ i ]，那么我们放入第 i 个银行以后，用 j 代表我们抢的钱，从10000向前遍历，一直到money[ i ]，状态转移方程为 dp[ j ] = max(dp[ j ] , pos[ i ]*dp[ j -money[ i ] ])

从后向前遍历是为了不重复，一个银行抢一次。做过背包的无需赘述。

下面贴一下代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>

using namespace std;

int money[105];
double pos[105];
double dp[10005];
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        double total;
        memset(dp,0,sizeof(dp));
        scanf("%lf%d",&total,&n);
        total = 1-total;

        for(int i = 0; i < n; i++){
            scanf("%d",&money[i]);
            scanf("%lf",&pos[i]);
            pos[i] = 1-pos[i];
        }

        dp[0] = 1;
        for(int i = 0; i < n; i++){//!!放入的序号

            for(int j = 10002; j >= money[i]; j--){
                dp[j] = max(dp[j],pos[i]*dp[j-money[i]]);
            }
        }
        for(int i = 10002; i >= 0; i--){
            if(dp[i] >= total){
                cout << i << endl;
                break;
            }
        }
    }
}