概率dp入门12/14

概率DP主要用于求解期望、概率等题目。

转移方程有时候比较灵活。

一般求概率是正推，求期望是逆推。通过题目可以体会到这点。

没有老师给题了只能自己去poj上补题

poj  2096 Collecint Bugs

Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 2634 Accepted: 1284Case 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：

有n类bug和s个子系统，bug数量不限，且每天只能发现一个bug，

要求的是当在s个子系统中发现n类bug时所需要天数的期望（平均天数）。

分析：

先确定状态，假设dp[i][j]表示已经在j个子系统中发现i类bug时所用天数的期望，明显dp[n][s] = 0。然后推导状态之间的转移，依据dp[i][j]的含义我们不难发现，下一天发现bug的情况只可能是以下四种情况：

1、在新的子系统中发现新的bug，即dp[i+1][j+1]

3、在已经发现过bug的子系统中发现新的bug，即dp[i+1][j]

2、在新的子系统中发现已经发现过的bug，即dp[i][j+1]

4、在已经发现过bug的子系统中发现已经发现过的bug，即dp[i+1][j+1]

同样，不难得出上述四种情况对应的概率分别为：p1 = (n-i)*(s-j) / (n*s)，p2 = (n-i)*(j) / (n*s)，p3 = i*(s-j) / (n*s)，p4 = i*j / (n*s)。

综上，我们的状态转移方程，同样根据期望的定义，dp[i][j] = p1*dp[i+1][j+1] + p2*dp[i+1][j] + p3*dp[i][j+1] + p4*dp[i][j] + 1，移项合并一下，dp[i][j] = (p1*dp[i+1][j+1] + p2*dp[i+1][j] + p3*dp[i][j+1] + 1) / (1-p4)。这样一来，我们要求的答案就是dp[0][0]。

源代码：

[cpp] view plain copy

1. #include <cstdio>  
2. #include <cstring>  
3.   
4. using namespace std;  
5.   
6. const int MAXN = 1005;  
7. double dp[MAXN][MAXN];  
8.   
9. int main()  
10. {//freopen("sample.txt", "r", stdin);  
11.     int n, s;  
12.     while(~scanf("%d%d", &n, &s))  
13.     {  
14.         double p1, p2, p3, p4;  
15.         memset(dp, 0, sizeof(dp));  
16.         for(int i=n; i>=0; --i)  
17.             for(int j=s; j>=0; --j)  
18.             {  
19.                 if(i==n && j==s)  
20.                     continue;  
21.                 p1 = 1.0*(n-i)*(s-j) / (n*s);  
22.                 p2 = 1.0*(n-i)*j     / (n*s);  
23.                 p3 = 1.0*i*(s-j)     / (n*s);  
24.                 p4 = 1.0*i*j         / (n*s);  
25.                 dp[i][j] = (p1*dp[i+1][j+1] + p2*dp[i+1][j]  // 状态转移  
26.                          + p3*dp[i][j+1]    + 1)  
27.                          / (1-p4);  
28.             }  
29.         printf("%.4f\n", dp[0][0]);  
30.     }  
31.     return 0;  
32. }