POJ-2096 Collecting Bugs （概率DP）

Collecting Bugs

http://poj.org/problem?id=2096
Time Limit: 10000MS Memory Limit: 64000K   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

完全是英语阅读题，n和s都不着重标注一下...

题目大意：有n种bug和s个子部件，每天只能找到一个bug，求集齐n种bug，且每个子部件上至少有一个bug的期望天数？

按照以前的观点，应该是设dp[i][j]表示在j个子部件中找到i种bug的期望天数，然后进行转移，但是题解都是倒推，不是很理解

【貌似是由于以前是无效状态，但对于后面的某种状态会有效，所以无法顺推。（而且好像这样算的话，由于仍然是n*s-i*j做分母，则目标状态一定算不出来）】

设dp[i][j]表示在j个子部件中找到i种bug时离最终状态的期望的天数，则有dp[n][s]=0

则dp[i][j]可由转化为下列四种状态：

①dp[i][j]：表示在已有bug的j个子部件中找到已发现过的bug，概率为：(i/n)*(j/s)

②dp[i+1][j]：表示在已有bug的j个子部件中找到一个新的bug，概率为：((n-i)/n)*(j/s)

③dp[i][j+1]：表示在一个新的子部件中找到一个已发现过的bug，概率为：(i/n)*((s-j)/s)

④dp[i+1][j+1]：表示在一个新的子部件中找到一个新的bug，概率为：((n-i)/n)*((s-j)/s)

则状态转移方程为：dp[i][j]=dp[i][j]*(i/n)*(j/s)+dp[i+1][j]*((n-i)/n)*(j/s)+dp[i][j+1]*(i/n)*((s-j)/s)+dp[i+1][j+1]*((n-i)/n)*((s-j)/s)+1;

将dp[i][j]*(i/n)*(j/s)移至左边后解得：dp[i][j]=(dp[i+1][j]*((n-i)/n)*(j/s)+dp[i][j+1]*(i/n)*((s-j)/s)+dp[i+1][j+1]*((n-i)/n)*((s-j)/s)+1)*(n*s)/(n*s-i*j);

整理得：dp[i][j]=(dp[i+1][j]*(n-i)*j+dp[i][j+1]*i*(s-j)+dp[i+1][j+1]*(n-i)*(s-j)+n*s)/(n*s-i*j);

#include <cstdio>#include <algorithm>using namespace std;int n,s;double dp[1005][1005];int main() {    while(2==scanf("%d%d",&n,&s)) {        dp[n][s]=0;        for(int i=n;i>=0;--i) {            for(int j=s;j>=0;--j) {                if(i!=n||j!=s) {                    dp[i][j]=(dp[i+1][j]*(n-i)*j+dp[i][j+1]*i*(s-j)+dp[i+1][j+1]*(n-i)*(s-j)+n*s)/(n*s-i*j);                }            }        }        printf("%.4lf\n",dp[0][0]);    }    return 0;}