poj 2096 Collecting Bugs 期望dp 简单题

Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 3941 Accepted: 1964Case 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

题目说分类1共n个类型，分类2共s个类型。

现在要搜集bug们，使分类1的每个类型都至少有1个，分类2也是。

每一个bug有且仅有一个对应的类型在分类1，有且仅有一个对应的类型在分类2。

每天找一个bug，bug在1和2中对应某个类型的概率均等。求满足要求的天数期望。（bug无限多）

首先，期望有线性性质：

E[i][j]表示目前搜集了i个分类1，j个分类2,此时到完成目标所需天数的期望。

容易得到：

E[i][j]=1+  i/n*j/s*E[i][j]+ (n-i)/n*j/s* E[i+1][j] + i/n*(s-j)/s*E[i][j+1]+(n-i)/n*(s-j)/s*E[i+1][j+1];

就是利用了上面那个性质。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI (4.0*atan(1.0))#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   ind<<1,le,mid#define rson    ind<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     secondusing namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn=1000+10    ;const int maxm=1000+10    ;double dp[maxn][maxm];int n,s;int main(){    while(~scanf("%d%d",&n,&s))    {        memset(dp,0,sizeof dp);        for(int i=n;i>=0;i--)        {            for(int j=s;j>=0;j--)            {                if(i==n&&j==s)  continue;                dp[i][j]=( n*s+(n-i)*j*dp[i+1][j]+i*(s-j)*dp[i][j+1]+(n-i)*(s-j)*dp[i+1][j+1])/(n*s-i*j);            }        }        printf("%.4f\n",dp[0][0]);    }    return 0;}