Poj 2096 Collecting Bugs (概率DP求期望)
来源:互联网 发布:微软系列软件 编辑:程序博客网 时间:2024/04/30 20:13
Description
Two companies, Macrosoft and Microhard are in tight competition. Microhard wants to decrease sales of one Macrosoft program. They hire Ivan to prove that the program in question is disgusting. However, Ivan has a complicated problem. This new program has s subcomponents, and finding bugs of all types in each subcomponent would take too long before the target could be reached. So Ivan and Microhard agreed to use a simpler criteria --- Ivan should find at least one bug in each subsystem and at least one bug of each category.
Macrosoft knows about these plans and it wants to estimate the time that is required for Ivan to call its program disgusting. It's important because the company releases a new version soon, so it can correct its plans and release it quicker. Nobody would be interested in Ivan's opinion about the reliability of the obsolete version.
A bug found in the program can be of any category with equal probability. Similarly, the bug can be found in any given subsystem with equal probability. Any particular bug cannot belong to two different categories or happen simultaneously in two different subsystems. The number of bugs in the program is almost infinite, so the probability of finding a new bug of some category in some subsystem does not reduce after finding any number of bugs of that category in that subsystem.
Find an average time (in days of Ivan's work) required to name the program disgusting.
Input
Output
Sample Input
1 2
Sample Output
3.0000
题目大意
一个软件有 s 个子系统,存在 n 种 bug。某人一天能找到一个 bug。问,在这个软件中找齐 n 种 bug,
并且每个子系统中至少包含一个 bug 的时间的期望值(单位:天)。注意:bug 是无限多的,每个 bug
属于任何一种 bug 的概率都是 1/n;出现在每个系统是等可能的,为 1/s。
做法分析
令 f[i][j] 表示已经找到了 i 种 bug,且 j 个子系统至少包含一个 bug,距离完成目标需要的时间的期望。
目标状态是 f[0][0]
再过一天找到一个 bug 可能是如下的情况:
1. 这个 bug 的种类是 已经找到的 并且 出现在 已经找到 bug 的子系统中
2. 这个 bug 的种类是 已经找到的 并且 出现在 没有找到 bug 的子系统中
3. 这个 bug 的种类是 没有找到的 并且 出现在 已经找到 bug 的子系统中
4. 这个 bug 的种类是 没有找到的 并且 出现在 没有找到 bug 的子系统中
经过简单的分析,不难得出如下递推过程:
f[i][j] = i/n*j/s*f[i][j]
+ i/n*(s-j)/s*f[i][j+1]
+ (n-i)/n*j/s*f[i+1][j]
+ (n-i)/n*(s-j)/s*f[i+1][j+1]
移项可得
(1-(i*j)/(n*s))f[i][j] = i/n*(s-j)/s*f[i][j+1]
+ (n-i)/n*j/s*f[i+1][j]
+ (n-i)/n*(s-j)/s*f[i+1][j+1]
逆向递推即可
#include <iostream>#include <stdio.h>#include <string.h>using namespace std;double f[1005][1005];int main(){ int n,s; while(scanf("%d%d",&n,&s)!=EOF) { memset(f,0,sizeof(f)); for(int i=n;i>=0;i--) { for(int j=s;j>=0;j--) { if(i==n&&j==s) continue; double p1=double(s-j)*i/n/s; double p2=double(n-i)*j/n/s; double p3=double(n-i)*(s-j)/n/s; double p0=1.0-double(i*j)/n/s; f[i][j]=p1*f[i][j+1]+p2*f[i+1][j]+p3*f[i+1][j+1]+1; f[i][j]/=p0; } } printf("%.4f\n",f[0][0]); } return 0;}
- POJ 2096 Collecting Bugs(概率DP求期望)
- poj 2096 Collecting Bugs 概率dp求期望
- POJ 2096 - Collecting Bugs(概率DP 求期望)
- poj 2096 Collecting Bugs 概率DP求期望(简单)
- POJ 2096 Collecting Bugs 概率dp 求期望 入门
- POJ 2096 Collecting Bugs(概率DP求期望)
- Poj 2096 Collecting Bugs (概率DP求期望)
- POJ 2096 Collecting Bugs(概率DP,求期望)
- poj 2096 Collecting Bugs【概率dp 逆向求期望】
- Collecting Bugs+POJ 2096+概率期望dp
- POJ 2096 Collecting Bugs 概率DP(期望)
- POJ 2096Collecting Bugs(概率期望dp)
- Poj 2096 Collecting Bugs (dp求期望)
- [ACM] poj 2096 Collecting Bugs (概率DP,期望)
- poj 2096 Collecting Bugs 概率dp(期望)
- Poj 2096 Collecting Bugs(概率(期望)+dp)
- poj 2096 Collecting Bugs 【概率DP】【逆向递推求期望】
- 【POJ 2096】Collecting Bugs(概率与期望+dp)
- HDU 5159 Card (概率求期望)
- 【转载】Tomcat 7.0.6X 启动慢并且遇到StackOverflowError的异常的解决办法
- Netty 缓存buffer介绍及使用
- eclipse 代码格式化
- Android EditText聚焦时hint消失的简单代码
- Poj 2096 Collecting Bugs (概率DP求期望)
- HDU 3853 LOOP (概率DP求期望)
- ZOJ 3822 Domination (三维概率DP)
- php 验证身份证有效性,根据国家标准GB 11643-1999 15位和18位通用
- poj 2318 TOYS (计算几何)
- Android Tools 在线更新SDK Manager
- poj 2398 Toy Storage (计算几何)
- poj3347 Kadj Squares (计算几何)
- Jenkins+maven+tomcat8自动构建部署配置