POJ2096 Collecting Bugs 解题报告【概率DP】
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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
解题报告
这道题是一个概率DP。
题目大意是这样的:
有s个系统,n种bug,我们需要多少天(也就是期望值)使得每个系统都发现bug。
我们设 dp[i][j] 表示找到i个bug,他们属于j个系统所产生的期望值。设需要m天 dp[0][0]=m,当i=n,j=s时,天数为0,则有 dp[n][s]=0 ,求dp[0][0]。
那么dp[i][j]就有四种状态了:
1.dp[i][j] 属于已经有的i个分类和j个系统 概率为 i/n*j/s ;
2.dp[i+1][j] 属于已有的j个系统和未有的分类 概率为 (1-i/n)*j/s;
3.dp[i][j+1] 属于已有的i个分类和未有的系统 概率为 (1-j/s)*i/n;
4.dp[i+1][j+1] 不属于已有的分类和系统 概率为 (1-i/n)*(1-j/s)
dp[i][j]是等于这四个中状态的概率之和,所以可以得出状态转移方程
dp[i][j] = p1*dp[i][j]+p2*[i+1][j]+p3*[i][j+1]+p4*[i+1][j+1];
将状态转移方程移位下可得:
dp[i][j]=((s-j)*i*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(s-j)*(n-i)*dp[i+1][j+1]+n*s)/(n*s-i*j);
代码如下:
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int N=1000;double dp[N+5][N+5];int n,s;int main(){ while(~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)continue; dp[i][j]=((s-j)*i*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(s-j)*(n-i)*dp[i+1][j+1]+n*s)/(n*s-i*j); } } printf("%.4f\n",dp[0][0]); } return 0;}
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