HDU2955 Robberies(01背包)

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23140    Accepted Submission(s): 8529

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：有一个强盗要去几个银行偷盗，他既想多偷点钱，又想尽量不被抓到。已知各个银行的金钱数和被抓的概率，以及强盗能容忍的最大被抓概率。求他最多能偷到多少钱？*/#include <bits/stdc++.h>#define ll long long#define PI acos(-1)#define M(n, m) memset(n, m, sizeof(n));const int INF = 1e9 + 7;const int maxn = 1e5;using namespace std;struct data{    int val;// 偷的钱数    double p;//被抓的概率} a[maxn];// dp 数组中存储不被抓的概率double p, dp[maxn];int sum, n;int main(){    int T;    cin >> T;    while (T --)    {        sum = 0;        cin >> p >> n;        for (int i = 0; i < n; i ++)        {            cin >> a[i].val >> a[i].p;            sum += a[i].val;//最大的钱数        }        M(dp, 0)        dp[0] = 1;//边界初始化        for (int i = 0; i < n; i ++)        {            for (int j = sum; j >= a[i].val; j --)                // 1 - a[i].p 不被抓的概率                dp[j] = max(dp[j], dp[j - a[i].val] * (1 - a[i].p));        }        // 从后往前遍历，当找到第一个不被抓的概率大于 1 - p的时候，就终止        for (int i = sum; i >= 0; i --)            if (dp[i] > 1 - p)            {                cout << i << endl;                break;            }    }    return 0;}