Poj 2096 Collecting Bugs (概率DP求期望)

最新推荐文章于 2021-03-30 14:32:48 发布

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本文介绍了一种通过概率计算预测软件被认定为“恶心”的平均时间的方法。具体来说，是计算在一个具有多个子系统和多种类型bug的软件中，找到所有类型的bug并在每个子系统中至少找到一个bug所需的平均天数。

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C - Collecting Bugs

Time Limit:10000MS Memory Limit:64000KB 64bit IO Format:%I64d & %I64u

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Description

Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exactly one bug in the program and adds information about it and its category into a spreadsheet. When he finds bugs in all bug categories, he calls the program disgusting, publishes this spreadsheet on his home page, and forgets completely about the program.
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

Input file contains two integer numbers, n and s (0 < n, s <= 1 000).

Output

Output the expectation of the Ivan's working days needed to call the program disgusting, accurate to 4 digits after the decimal point.

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;
}