poj2096——Collecting Bugs(概率DP)

本文探讨了一个关于软件缺陷收集的问题,通过概率DP方法解决了一个复杂的软件评估挑战。具体来说,问题涉及如何评估一个由多个子系统构成的软件,在所有子系统中至少发现一个属于各类别的缺陷所需的平均时间。

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Collecting Bugs
Time Limit: 10000MS Memory Limit: 64000K
Total Submissions: 1444 Accepted: 623
Case Time Limit: 2000MS Special Judge
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
Source

Northeastern Europe 2004, Northern Subregion

解析:

         这是本人概率DP的入门题,觉得这是个很神奇的东东。。。

        至于详细解释,参见这篇博客http://blog.youkuaiyun.com/morgan_xww/article/details/6774708

代码:

#include<cstdio>

int n,s;
double dp[1010][1010];

int main()
{
    freopen("poj2096.in","r",stdin);       //dp[i][j]=p1*dp[i][j]+p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i+1][j+1]
    freopen("poj2096.out","w",stdout);     //p1=(i*j)/(n*s)  p2=((n-i)*j)/(n*s)  p3=(i*(s-j))/(n*s)   p4=((n-i)*(s-j))/(n*s)
    scanf("%d%d",&n,&s);
    dp[n][s]=0.0;
    int ns=n*s;
    for(int i=n;i>=0;i--)
        for(int j=s;j>=0;j--)
        {
            if(n==i && s==j)continue;
            dp[i][j]=(ns+(n-i)*j*dp[i+1][j] + i*(s-j)*dp[i][j+1] + (n-i)*(s-j)*dp[i+1][j+1])/(ns-i*j);
        }
    printf("%.4lf",dp[0][0]);
    return 0;
}


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