HD 2955 Robberies(0-1背包)
来源:互联网 发布:视频字幕添加软件 编辑:程序博客网 时间:2024/06/08 11:56
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2955
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
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
题意: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;注意精度。
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;注意精度。
转移方程为:dp[j]=max(dp[j-v[i]]*p[i],dp[j]);
如下代码:
如下代码:
#include<cstdio>#include<cstring>#include<iostream>#define esp 1e-10#define M 10005using namespace std;int main(){int t;scanf("%d",&t);while(t--){double p[M],sp;//p为抢每个银行的安全概率int v[M];//memset(dp,0,sizeof(dp));double dp[M]={1.0};//dp为拿完钱后自己的安全概率 初始状态最安全,所以为1.0 int n,s=0;scanf("%lf%d",&sp, &n);sp=1-sp;//转换成安全概率 for(int i=0; i<n; ++i){scanf("%d%lf", &v[i],&p[i]);s+=v[i];p[i]=1-p[i];//转换成安全概率 }for(int i=0; i<n; ++i){for(int j=s; j>=v[i]; --j){dp[j]=max(dp[j-v[i]]*p[i],dp[j]);}}for(int i=s; i>=0; --i)if(dp[i]-sp>esp){printf("%d\n",i);break;}}return 0;}
0 0
- HD 2955 Robberies(0-1背包)
- hdu 2955 Robberies(0/1背包)
- hdu 2955 Robberies (0-1背包)
- HDU 2955 Robberies (0-1背包)
- HDOJ 2955Robberies(0 1 背包)
- hdu 2955 dp(0,1背包) Robberies
- HDOJ 2955 Robberies (0/1背包)
- hdu 2955 Robberies(0/1背包)
- hdu 2955 Robberies (0 1背包)
- HDU(2955)Robberies (0-1背包)
- HDU 2955(Robberies)0-1背包问题
- hdoj 2955 Robberies 【0 1背包】
- HDOJ 2955 Robberies 【0 1背包】
- HDU2955-Robberies(0-1背包)
- HUD 2955 Robberies [0-1背包的简单转化]
- HDU 2955 Robberies (想法题&0-1背包)
- HDU 2955 Robberies 0-1背包 浮点数处理
- HDU 2955 Robberies(0-1背包问题的简单变形)
- shell 脚本:使用过程中使得变量能够赋值
- 数据库连接的config.properties配置文件
- java基础知识汇总4
- 怎么把PDF中的文字提取出来
- C#基本功修炼日记
- HD 2955 Robberies(0-1背包)
- NavigationView 的使用
- 树莓派3 wifi设置
- 用static关键字修饰类
- js面向对象创建
- [Android基础]Android总结篇
- 删除已经配置的类库和移除CocoaPods
- java基础知识总结5
- JAVA LinkedList和ArrayList的使用及性能分析