poj 2096 Collecting Bugs （期望dp入门）
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Collecting Bugs
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
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题意：
一个软件有s个子系统，会产生n种bug
某人一天发现一个bug,这个bug属于一个子系统，属于一个分类
每个bug属于某个子系统的概率是1/s,属于某种分类的概率是1/n
问发现n种bug,每个子系统都发现bug的天数的期望。
思路：
求期望从后往前比较好想。
dp[i][j]表示已经找到了i个bug，这i个bug属于j个系统，到达最终状态dp[n][m]所需要的天数的期望。
那么我们要求的就是dp[0][0]了。
到达dp[i][j]这个状态后，花一天，可能到达四种状态：
1:dp[i][j],概率为 i/n*j/s;
2:dp[i][j+1],概率为 i/n*(s-j)/s;
3:dp[i+1][j],概率为 (n-i)/n*j/s;
4:dp[i+1][j+1],概率为 (n-i)/n*(s-j)/s;
那么有dp[i][j]=1+i/n*j/s*dp[i][j]+(n-i)/n*j/s*dp[i][j+1]+(n-i)/n*j/s*dp[i+1][j]+ (n-i)/n*(s-j)/s*dp[i+1][j+1];
化简之后便可以得到方程了。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 400005
#define OO (1<<31)-1
#define mod 90003
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int n,s;
double dp[maxn][maxn];
int main()
{
int i,j,t;
while(~scanf("%d%d",&n,&s))
{
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+i*(s-j)*dp[i][j+1]/n/s+(n-i)*j*dp[i+1][j]/n/s+(n-i)*(s-j)*dp[i+1][j+1]/n/s)/(1-i*j*1.0/n/s);
}
}
printf("%.4f\n",dp[0][0]);
}
return 0;
}