poj2096——概率DP

Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 5371 Accepted: 2656Case 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

Source

Northeastern Europe 2004, Northern Subregion

KB大神说得对，正推求概率，逆推求期望~

题意：（题意看题目确实比较难道，n和s都要找半天才能找到）   一个软件有s个子系统，会产生n种bug   某人一天发现一个bug,这个bug属于一个子系统，属于一个分类   每个bug属于某个子系统的概率是1/s,属于某种分类的概率是1/n   问发现n种bug,每个子系统都发现bug的天数的期望。

转的题解：

1. 解法： 
2.     dp[i][j]表示已经找到i种bug，并存在于j个子系统中，要达到目标状态的天数的期望。 
3.     显然，dp[n][s]=0，因为已经达到目标了。而dp[0][0]就是我们要求的答案。 
4.     dp[i][j]状态可以转化成以下四种： 
5.         dp[i][j]    发现一个bug属于已经找到的i种bug和j个子系统中 
6.         dp[i+1][j]  发现一个bug属于新的一种bug，但属于已经找到的j种子系统 
7.         dp[i][j+1]  发现一个bug属于已经找到的i种bug，但属于新的子系统 
8.         dp[i+1][j+1]发现一个bug属于新的一种bug和新的一个子系统 
9.     以上四种的概率分别为： 
10.     p1 =     i*j / (n*s) 
11.     p2 = (n-i)*j / (n*s) 
12.     p3 = i*(s-j) / (n*s) 
13.     p4 = (n-i)*(s-j) / (n*s) 
14.     又有：期望可以分解成多个子期望的加权和，权为子期望发生的概率，即 E(aA+bB+...) = aE(A) + bE(B) +... 
15.     所以： 
16.     dp[i,j] = p1*dp[i,j] + p2*dp[i+1,j] + p3*dp[i,j+1] + p4*dp[i+1,j+1] + 1; 
17.     整理得： 
18.     dp[i,j] = ( 1 + p2*dp[i+1,j] + p3*dp[i,j+1] + p4*dp[i+1,j+1] )/( 1-p1 ) 
19.             = ( n*s + (n-i)*j*dp[i+1,j] + i*(s-j)*dp[i,j+1] + (n-i)*(s-j)*dp[i+1,j+1] )/( n*s - i*j ) 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <stack>
using namespace std;
typedef long long ll;
#define PI 3.1415926535897932
#define E 2.718281828459045
#define INF 0x3f3f3f3f
#define mod 100000007


const int M=1005;
int n,m;
int cnt;
int sx,sy,sz;
int mp[1000][1000];
int pa[M*10],rankk[M];
int head[M*6],vis[M*100];
int dis[M*100];
ll prime[M*1000];
bool isprime[M*1000];
int lowcost[M],closet[M];
char st1[5050],st2[5050];
int len[M*6];
typedef pair<int ,int> ac;
//vector<int> g[M*10];
double dp[M][M];
int has[10500];
int month[13]= {0,31,59,90,120,151,181,212,243,273,304,334,0};
int dir[8][2]= {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,1},{-1,-1}};


void getpri()
{
    ll i;
    int j;
    cnt=0;
    memset(isprime,false,sizeof(isprime));
    for(i=2; i<1000000LL; i++)
    {
        if(!isprime[i])prime[cnt++]=i;
        for(j=0; j<cnt&&prime[j]*i<1000000LL; j++)
        {
            isprime[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
}
struct node
{
    int v,w;
    node(int vv,int ww)
    {
        v=vv;
        w=ww;
    }
};
vector<int> g[M*100];
string str[1000];
//int dp[10000];
//int a[1000],b[1000];
int ans;


int main()
{
    int i,j,k,t;


    while(~scanf("%d%d",&n,&m))//m==s
    {
        for(i=n; i>=0; i--)
            for(j=m; j>=0; j--)
            {
                if(i==n&&j==m)dp[n][m]=0.0;
                else
                {
                    double p1=j*(n-i)*1.0;
                    double p2=(m-j)*i*1.0;
                    double p3=(m-j)*(n-i)*1.0;
                    dp[i][j]=(p1*dp[i+1][j]+p2*dp[i][j+1]+p3*dp[i+1][j+1]+n*m)/(n*m-i*j);
                }
            }
            printf("%.4f\n",dp[0][0]);
        return 0;
    }
}