poj 2096(概率dp求期望)

题目连接：http://poj.org/problem?id=2096

Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 2623 Accepted: 1275Case 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

思路：题意扯的比较复杂，但是思路比较好想，算作一道简单但又经典的概率dp了

有n种bug,和s个子系统，花费一天能找出一个bug来，这个bug只能属于n种bug中的一种，同时也只能属于s个子系统中一个；概率分别是 1/n,1/s；

那么问题来了：现在要将n种bug都找出来，并且s个子系统中至少有一个Bug;(bug数量无限大)，求天数的期望

用dp[i][j]表示 现在找出了i种bug,位于j个不同子系统上所要的天数期望；

那么容易知道dp[n][s]==0,要求的就是dp[0][0];

状态转移方程式： dp[i][j]=p1*dp[i][j]+p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i+1][j+2];

#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <algorithm>using namespace std;int n,s;double dp[1100][1100];void solve(){    dp[n][s]=0;    for(int i=n;i>=1;i--)        for(int j=s;j>=1;j--)        {           if(i==n&&j==s)continue;           double p1=1.0*(n-i)/n;           double p2=1.0*i/n;           double p3=1.0*(s-j)/s;           double p4=1.0*j/s;           dp[i][j]=(p2*p4+p1*p4*(dp[i+1][j]+1)+p2*p3*(dp[i][j+1]+1)+p1*p3*(dp[i+1][j+1]+1))/(1-p2*p4);           //printf("%d %d %.4lf\n",i,j,dp[i][j]);        }}int main(){   while(scanf("%d%d",&n,&s)!=EOF)   {       solve();       printf("%.4lf\n",dp[1][1]+1);   }   return 0;}