Robberies
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Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 23107 Accepted Submission(s): 8516
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.
![](http://acm.hdu.edu.cn/data/images/con211-1010-1.jpg)
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.
![](http://acm.hdu.edu.cn/data/images/con211-1010-1.jpg)
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
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
Roy想要抢劫银行,每家银行多有一定的金额和被抓到的概率,知道Roy被抓的最大概率P,求Roy在被抓的情况下,抢劫最多。
分析:被抓概率可以转换成安全概率,Roy的安全概率大于1-P时都是安全的。抢劫的金额为0时,肯定是安全的,所以d[0]=1;其他金额初始为最危险的所以概率全为0;
注意:不要误以为精度只有两位。
/*分类:01背包 来源:hdu Robberies思路:此题关键是找到背包的容量和价值We are giants.create by Lee_SD on 2017/4/17*/#include<stdio.h>#include<math.h>#include<string.h>#define MAX(a,b) (a>b?a:b)int main(){int T, i, j, n, M[110];float P, p[110], dp[10010];//用float 类型就行了, 用double 类型会导致数据缺失。scanf("%d",&T);while(T--){memset(dp,0,sizeof(dp));//要对数组进行初始化int sum = 0;scanf("%f%d",&P,&n);P = 1 - P;//成功逃走的概率for(i = 1; i <= n; i++){scanf("%d%f",&M[i],&p[i]);sum += M[i];//算出银行总钱数p[i] = (1 - p[i]);}dp[0] = 1;//未抢劫一分钱那么逃走的概率就为1for(i = 1; i <= n; i++){for(j = sum; j >= M[i]; j--){dp[j] = MAX(dp[j],dp[j - M[i]] * p[i]);//这里的每次成功的概率需要进行乘法运算,因为是两次成功的概率所以是乘法}}for(i = sum; i >= 0; i--){if(dp[i] >= P){break;//找出可以成功逃走并且逃走概率不超过警戒值}}printf("%d\n",i);}return 0;}
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