poj 2096 Collecting Bugs 概率dp求期望

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Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 2105 Accepted: 1006Case 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

题意：有一款软件有s个子系统，和n种bug。一个人每天只能发现一个bug，可以知道bug的种类和所属于的子系统，求找到n种bug并且每个系统至少有一个bug所需天数的期望。

思路：设dp[i][j]为已经发现i种bug，且存在于j个子系统中，要到达目标状态所需天数的期望。 由于每发现一个bug，属于某种bug的概率为1/n,属于某个子系统的概率为1/s，那么由dp[i][j]的状态我们有以下4种转移：

（1）dp[i][j] ------>dp[i][j] ,转移概率p1=( i / n ) * ( j / s );

（2）dp[i][j] ------>dp[i+1][j],转移概率p2=( ( n - i ) / i ) * ( j / s);

（3）dp[i][j] ------>dp[i][j+1],转移概率p3=( i / n) * ( ( s - j ) / s );

（4）dp[i][j] ------>dp[i+1][j+1],转移概率p4=( ( n - i ) / n ) * ( ( s - j ) / s );

又：期望可以分解成多个子期望的加权和，权因子为子期望发生的概率。

所以dp[i][j]=p1 * ( dp[i][j] + 1) + p2 * ( dp[i+1][j] + 1 ) + p3 * ( dp[i][j+1] + 1) + p4 * ( dp[i+1][j+1] + 1 ),

化简得dp[i][j] : dp[i][j] =(1 + p2 * dp[i+1][j] + p3 * dp[i][j+1] + p4 * dp[i+1][j+1] ) / ( 1 - p1) , 那么dp[0][0]即为最后答案。详见答案：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=1000+100;double dp[MAXN][MAXN];int n,m;int main(){    freopen("text.txt","r",stdin);    while(~scanf("%d%d",&n,&m))    {        dp[n][m]=0.0;        for(int i=n;i>=0;i--)            for(int j=m;j>=0;j--)            {                if(i==n && j==m)                    continue;                 dp[i][j]=(dp[i][j+1]*i*(m-j)/(1.0*n*m)+dp[i+1][j]*(n-i)*j/(1.0*n*m)+dp[i+1][j+1]*(n-i)*(m-j)/(1.0*n*m)+1.0)/(1-i*j*1.0/(1.0*n*m));            }        printf("%.4f\n",dp[0][0]);    }    return 0;}