[LightOJ 1079] Just another Robbery

Posted 2020-09-28

 Just another Robbery

As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermione and Ron have decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives an integer M_j (0 < M_j ≤ 100) and a real number P_j . Bank j contains M_j millions, and the probability of getting caught from robbing it is P_j. A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output

For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

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

Case 1: 2

Case 2: 4

Case 3: 6

Hint

For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (1 - 0.02) * .03 = 0.0494 which is greater than the given probability (0.04). That‘s why he has only option, just to rob rank 2.

 

吐槽一句,好久没见过HP的题目了,然而现在HP被改造成了反派......

题目的意思是,给出几个银行,每个银行有一定数量的钱,当然在这个银行作案也有一定几率被抓.HP想要抢劫一些银行,不仅要使被抓的概率小于一个固定值,还要抢到尽可能多的钱.

 这比较容易看出来是一个01背包问题,既有条件,也有价值.

我们设E[i]为抢到数量为i的钱的逃跑成功的概率(当然,你也可以设被抓的概率),则根据乘法原理,我们得出一个转移方程:E[j]=max(E[j],E[j-M[i]]*P[j]).

最后只要从allM到0扫一遍,如果E[i]>那个给出的固定值,则就输出i就行了.

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 int n,M[105];
 7 double EXP,P[105],E[10005];
 8 int main(){
 9     int T; scanf("%d",&T);
10     for (int Ts=1; Ts<=T; Ts++){
11         scanf("%lf%d",&EXP,&n),EXP=1-EXP; int allM=0;
12         for (int i=1; i<=n; i++) scanf("%d%lf",&M[i],&P[i]),allM+=M[i],P[i]=1-P[i];
13         memset(E,0,sizeof E),E[0]=1;
14         for (int i=1; i<=n; i++)
15             for (int j=allM; j>=M[i]; j--) E[j]=max(E[j],E[j-M[i]]*P[i]);
16         for (int i=allM; i>=0; i--) if (E[i]>EXP){printf("Case %d: %d\n",Ts,i); break;}
17     }
18     return 0;
19 }

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