POJ 2096 Collecting Bugs 期望dp

题目链接：

http://poj.org/problem?id=2096

Collecting Bugs

Time Limit: 10000MS
Memory Limit: 64000K #### 问题描述 > 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. #### 输入 > Input file contains two integer numbers, n and s (0 < n, s <= 1 000). #### 输出 > Output the expectation of the Ivan's working days needed to call the program disgusting, accurate to 4 digits after the decimal point. ####样例输入 > 1 2

样例输出

3.0000

题意

一个人每天可以发现一个bug，这个bug属于n个类别的任意一个（等概率），位于s个子系统的任意一个（也是等概率，且和类别独立），问这个人发现n类bug，且每个子系统至少有1个bug被发现的期望次数是多少。

题解

我们定义dp[i][j]为已经发现了i类bug，位于j个子系统的，还需多少次才能发现n类bug位于s个子系统，且每个子系统至少一个bug的期望次数。
对于发现的一个bug，有以下四种情况：
1、属于已经发现的类的bug，属于已经发现的子系统：dp[i][j],p1=i/n*j/s
2、不属于已经发现的bug，属于已经发现的子系统：dp[i+1][j],p2=(n-i)/n*j/s
3、属于已经发现的bug，不属于已经发现的子系统：dp[i][j+1],p3=i/n*(s-j)/s
4、不属于已经发现的bug，不属于已经发现的子系统：dp[i+1][j+1],p4=(n-i)/n*(s-j)/s
然后再考虑下转移过来的花费：1天，既：
dp[i][j]=p1dp[i][j]+p2dp[i+1][j]+p3dp[i][j+1]+p4dp[i+1][j+1]+1;
移项得：
dp[i][j]=(p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i+1][j+1]+1)/(1-p1).

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1010;

double dp[maxn][maxn];
int n,s;

int main() {
    while(scf("%d%d",&n,&s)==2&&n){
        for(int i=n;i>=0;i--){
            for(int j=s;j>=0;j--){
                if(i==n&&j==s) dp[n][s]=0;
                else{
                    double p1=1.0*i*j/(n*s);
                    double p2=1.0*(n-i)*j/(n*s);
                    double p3=1.0*i*(s-j)/(n*s);
                    double p4=1.0*(n-i)*(s-j)/(n*s);
                    dp[i][j]=(p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i+1][j+1]+1)/(1-p1);
                }
            }
        }
        prf("%.4f\n",dp[0][0]);
    }
    return 0;
}

//end-----------------------------------------------------------------------