Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21310 Accepted Submission(s): 7885
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
Source
题意：
一人去抢银行，n个银行，在每个银行能够得到的钱mj,被抓住的概率pj，抢所有银行被抓住的概率p，问在总概率不超过p时得到的最大钱数。
代码：
1 //由于存在概率的乘法，用普通的01背包肯定不行，可以以总钱数作为背包的容量，求不被抓到的最大概率，最后for语句，钱数递减找到第一个符合的概率即可。
2 //注意初始化背包时f[0]=1,其他的是0；被抓的概率是1减去不被抓的概率。
3 #include<iostream>
4 #include<cstdio>
5 using namespace std;
6 int t,n;
7 double p,pj[105];
8 int mj[105];
9 double f[10004];
10 int main()
11 {
12 scanf("%d",&t);
13 while(t--)
14 {
15 int sum=0;
16 for(int i=1;i<=10000;i++)
17 f[i]=0;
18 f[0]=1;
19 scanf("%lf%d",&p,&n);
20 for(int i=1;i<=n;i++)
21 {
22 scanf("%d%lf",&mj[i],&pj[i]);
23 sum+=mj[i];
24 pj[i]=1-pj[i];
25 }
26 for(int i=1;i<=n;i++)
27 {
28 for(int k=sum;k>=mj[i];k--)
29 {
30 f[k]=max(f[k],f[k-mj[i]]*pj[i]);
31 }
32 }
33 for(int i=sum;i>=0;i--)
34 {
35 if(1-f[i]<=p)
36 {
37 printf("%d\n",i);
38 break;
39 }
40 }
41 }
42 return 0;
43 }