hdu - 2955 - Robberies

Robberies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 342 Accepted Submission(s): 154

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

 

唉，这道题以为是简单的01背包，结果一直WA，找了好久的错，终于发现事情不是想象的那样，

我之前一直是将被抓概率相加，只要小于上限被抓概率就好，但是，每个银行抢劫都是独立事件，

也就是说如果有两家银行A和B，被抓概率分别为0.3和0.4，那么理论有四种结果，被A抓没被B抓，

被B抓没被A抓，被A，B抓，没有被抓，而我们要的正是没有被任何银行抓的概率！所以反求其

生存率！

#include<iostream>#include<algorithm>using namespace std;typedef struct{    int money;    double risk;}Bank;double escapeRisk[10005];int main(){    int testCase;    cin >> testCase;    while(testCase--)    {        int bankNumber,maxMoney = 0;        double bankRisk;        Bank bank[105];        memset(escapeRisk,0,sizeof(escapeRisk));        escapeRisk[0] = 1.0;        cin >> bankRisk >> bankNumber;        for(int i = 1; i <= bankNumber; i++)        {            cin >> bank[i].money >> bank[i].risk;            maxMoney += bank[i].money;        }        for(int num = 1; num <= bankNumber; num++)        {            for(int limit = maxMoney; limit >= bank[num].money; limit--)            {                escapeRisk[limit] = max(escapeRisk[limit],escapeRisk[limit-bank[num].money]*(1-bank[num].risk));            }        }        for(int i = maxMoney; i >= 0; i--)        {            if((1-bankRisk) <= escapeRisk[i])             {                cout << i << endl;                break;            }        }    }    return 0;}