Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26790 Accepted Submission(s): 9863
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.
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 .
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.
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
题目大意：给出一个整数N和一个概率P，一个小偷抢劫N间银行，
抢劫第i间银行都有一个被抓概率p[i]，问在被抓概率不超过P的情况下，小偷最多能抢到多少钱。
分析：01背包。刚开始用概率*100做背包容量，但概率不止两位小数，所以需要转换思路，
用所有银行总资产做背包容量。dp[i][j]为决定前i间银行是否抢劫时，抢劫的钱数刚好 为j时，
小偷被抓的最小概率，未抢劫时被抓的概率为0，即dp[0]=0；注意被抓概率用乘积，
小偷被抓的最小概率，未抢劫时被抓的概率为0，即dp[0]=0；注意被抓概率用乘积，
那么滚动数组的状态转移方程为：dp[j]=min(dp[j],1-(1-dp[j-money[i]])*(1-p[i]))。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 int money[110]; 6 double dp[10100],p[110]; 7 int main() 8 { 9 int T,N; 10 double P; 11 scanf("%d",&T); 12 while(T--) 13 { 14 int M=0; 15 scanf("%lf%d",&P,&N); 16 for(int i=1;i<=N;i++) 17 { 18 scanf("%d%lf",&money[i],&p[i]); 19 M+=money[i]; 20 } 21 dp[0]=0;//未抢劫时被抓的概率为0 22 for(int i=1;i<=M;i++)dp[i]=2.0;//初始化 23 for(int i=1;i<=N;i++) 24 for(int j=M;j>=money[i];j--) 25 if(1-(1-dp[j-money[i]])*(1-p[i])<=P) 26 dp[j]=min(dp[j],1-(1-dp[j-money[i]])*(1-p[i])); 27 //dp[i-1][j]为不抢第i间的概率 28 //1-(1-dp[j-money[i]])*(1-p[i])为抢第i间的概率 29 int ans=0; 30 for(int i=M;i>=0;i--) 31 { 32 if(dp[i]<=P&&dp[i]>0){ans=i;break;} 33 } 34 printf("%d\n",ans); 35 } 36 return 0; 37 }
