poj 2096 Collecting Bugs 【概率DP】【逆向递推求期望】

Collecting Bugs

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 3523		Accepted: 1740
Case 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

题意：一个软件有S个系统。但会产生N种bug。一个人一天能够发现一个bug,这个bug既属于某一个系统。又属于某一个分类。每一个bug属于某个系统的概率是1/S,属于某种分类的概率是1/N。如今问你发现N种bug且S个系统都发现bug的天数的期望。

思路：用dp[i][j]表示发现i种bug且j个系统都发现bug的天数的期望。能够得到4种状态

1。dp[i][j]—— 新bug  既属于已发现bug的种类。又属于已经发现bug的系统。则有概率 (i / S) * (j / N)

2。dp[i+1][j]—— 新bug  不属于已发现bug的种类，属于已经发现bug的系统。则有概率 (1 - i / N) * (j / S)

3。dp[i][j+1]—— 新bug  属于已发现bug的种类，不属于已经发现bug的系统。则有概率 (i / N) * (1 - j / S)

4，dp[i+1][j+1]—— 新bug 既不属于已经发现bug的种类，也不属于已经发现bug的系统。

则有概率 (1 - i / N) * (1 - j / S)

求期望倒着递推，由dp[N][S] = 0 退出 dp[0][0]就可以。注意推导时 全是浮点型。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
double dp[1010][1010];
int main()
{
    int N, S;
    while(scanf("%d%d", &N, &S) != EOF)
    {
        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;
                double x = i, y = j;
                double p1 = dp[i+1][j] * (y / S) * (1 - x / N);
                double p2 = dp[i][j+1] * (x / N) * (1 - y / S);
                double p3 = dp[i+1][j+1] * (1 - y / S) * (1 - x / N);
                double p0 = 1 - (x / N) * (y / S);
                dp[i][j] = (p1 + p2 + p3 + 1) / p0;
            }
        }
        printf("%.4lf\n", dp[0][0]);
    }
    return 0;
}