POJ 2096 Collecting Bugs (概率DP）

Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 4038 Accepted: 2013Case 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

题目链接：点击打开题目链接

题意：一个程序有s个子系统，n种bug，一天发现一个bug，这个bug出现在每个子系统的概率相同且不会同时属于两个子系统，同理，是每种bug的概率相同。求发现每个子系统至少一个bug，每种bug至少一个的天数期望。

思路：dp[i][j]表示i个子系统出现bug，且共发现j种bug时达到目标状态的天数期望。dp[s][n]=0;

             状态转移方程：

             接下来发现的bug有四种可能：

             属于一个新的子系统，是一种新bug，概率p1=（1-i/s)*(1-j/n);

             属于一个新的子系统，不是一种新bug，概率p2=（1-i/s)*j/n;

             不属于一个新的子系统，是一种新bug，概率p3=i/s*(1-j/n);

             不属于一个新的子系统，不是一种新bug，概率p4=/s*j/n;

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

            化简得dp[i][j]=(1+p1*dp[i+1][j+1]+p2*dp[i+1][j]+p3*dp[i][j+1])/(1-p4);

代码：

#include<stdio.h>#include<string.h>double dp[1010][1010];int main(){    int n,s,i,j;    while(scanf("%d%d",&n,&s)!=EOF)    {    dp[n][s]=0;    for(i=n;i>=0;i--)    for(j=s;j>=0;j--)    {        if(i==n&&j==s) continue;        dp[i][j]=1.0*((n-i)*(s-j)*dp[i+1][j+1]+i*(s-j)*dp[i][j+1]+n*s+(n-i)*j*dp[i+1][j])/(n*s-i*j);//避免损失精度分子分母同乘以n*s    }    printf("%.4lf\n",dp[0][0]);    }}