poj2096 Collecting Bugs

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。

求解：
         dp[i][j]表示已经找到i种bug,j个系统的bug，达到目标状态的天数的期望
         dp[n][s]=0;要求的答案是dp[0][0];
         dp[i][j]可以转化成以下四种状态:
              dp[i][j],发现一个bug属于已经有的i个分类和j个系统。概率为(i/n)*(j/s);
              dp[i][j+1],发现一个bug属于已有的分类，不属于已有的系统.概率为 (i/n)*(1-j/s);
              dp[i+1][j],发现一个bug属于已有的系统，不属于已有的分类,概率为 (1-i/n)*(j/s);
              dp[i+1][j+1],发现一个bug不属于已有的系统，不属于已有的分类,概率为 (1-i/n)*(1-j/s);

        整理便得到转移方程

http://www.cnblogs.com/zhj5chengfeng/archive/2013/03/02/2939597.html

http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710621.html

补充一下，为啥要+1，因为转移到四种情况，需要贡献1，

所以总体可以这样理解，第一种情况贡献了它的(i/n)*(j/s),第二种情况贡献了它的 (i/n)*(1-j/s),以此类推，

于是，dp[0][0]就是最终答案

#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;double dp[1010][1010];int main(){int n,s;cin>>n>>s;for(int i=n;i>-1;i--)for(int j=s;j>-1;j--){if(i==n&&j==s)continue;double t1=double(i)/n,t2=double(j)/s;dp[i][j]=dp[i+1][j]*(1-t1)*t2+dp[i][j+1]*t1*(1-t2)+dp[i+1][j+1]*(1-t1)*(1-t2)+1;dp[i][j]/=(1-t1*t2);}printf("%.4f\n",dp[0][0]);}

Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 3673 Accepted: 1827Case 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