【hdu2955】Robberies——01背包

题目：

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18298    Accepted Submission(s): 6769

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

 

Source

IDI Open 2009

描述：给定抢银行被抓的概率和银行数，再分别给出每个银行能抢到的钱数以及被抓的概率，求不被抓的情况下能抢到的最大钱数。

题解：首先能肯定的是肯定不能用概率做下标去dp钱数，那么第一可以考虑dp[i]表示抢i百万元钱时不被抓的概率，最后从sum开始找到第一个小于总的不被抓的概率的钱数i输出。

代码：

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <iostream>using namespace std;const int maxn = 105;int  v[maxn];double dp[10005];int main(){//freopen("input.txt", "r", stdin);int T, N;double P, w[maxn] = { 0.0 };scanf("%d", &T);while (T--){scanf("%lf%d", &P, &N);int sum = 0;for (int i = 0; i < N; i++){cin >> v[i] >> w[i];sum += v[i];}memset(dp, 0, sizeof(dp));dp[0] = 1;for (int i = 0; i < N; i++)for (int j = sum; j >= v[i]; j--)dp[j] = max(dp[j], dp[j - v[i]] * (1 - w[i]));for (int i = sum; i >= 0; i--)if (dp[i] > 1-P){cout << i << endl;break;}}return 0;}