poj 2096 Collecting Bugs 概率DP求期望（简单）

题目链接：http://poj.org/problem?id=2096

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

题意：   s个子系统，n种bug，每天找一个bug，求找到所有n种bug且每个子系统至少有一个bug的天数的期望 。

思路：
   dp[i][j]表示找到i种bug并且在j个子系统中的天数的期望，dp[n][s]=0；倒着推，dp[0][0]就是答案。
  dp[i][j]状态可以转化成以下四种：
  dp[i][j]发现新的bug是已经找到的i种bug和j个子系统中
  dp[i+1][j]发现新的bug是新的一种bug，但属于已经找到的j个子系统中
  dp[i[j+1]发现新的bug是已经找到的i种bug，但属于新的一个子系统中
  dp[i+1][j+1]发现新的bug是新的一种bug，也是新的一个子系统中
  与其对应的四种概率：
  p1=i*j/(n*s);
  p2=(n-i)*j/(n*s);
  p3=i*(s-j)/(n*s);
  p4=((n-i)*(s-j))/(n*s);
  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) +...
  所以：
  dp[i,j] = p1*dp[i,j] + p2*dp[i+1,j] + p3*dp[i,j+1] + p4*dp[i+1,j+1] + 1;  //+1表示是新的一天
  整理得：
  dp[i,j] = ( 1 + p2*dp[i+1,j] + p3*dp[i,j+1] + p4*dp[i+1,j+1] )/( 1-p1 ) = ( 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 )

 代码：

#include <algorithm>#include <cstdlib>#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <cctype>#include <cmath>#include <stack>#include <queue>#include <list>#include <map>#include <set>using namespace std;#define min2(x, y)     min(x, y)#define max2(x, y)     max(x, y)#define min3(x, y, z)  min(x, min(y, z))#define max3(x, y, z)  max3(x, max(y, z))#define clr(x, y)      memset(x, y, sizeof(x))#define fr(i,n)        for(int i = 0; i < n; i++)#define fr1(i,n)       for(int i = 1; i <= n; i++)#define upfr(i,j,n)    for(int i = j; i <= n; i++)#define dowfr(i,j,n)   for(int i = n; i >= j; i--)#define scf(n)         scanf("%d", &n)#define scf2(n,m)      scanf("%d %d",&n,&m)#define scf3(n,m,p)    scanf("%d %d %d",&n,&m,&p)#define ptf(n)         printf("%d",n)#define ptf64(n)       printf("%I64d",n)#define ptfs(s)        printf("%s",s)#define ptln()         printf("\n")#define ptk()          printf(" ")#define ptc(c)         printf("%c",c)#define srt(a,n)       sort(a,n)#define LL long long#define pi acos(-1.0)#define inf 1 << 31-1#define eps 0.00001#define maxn 1005#define mod 10000007double dp[maxn][maxn];int main(){// freopen("in.txt","r",stdin);int n,s;while(scf2(n, s) == 2){clr(dp, 0);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]=(n*s + i*(s-j)*dp[i][j+1] + (n-i)*j*dp[i+1][j] + (n-i)*(s-j)*dp[i+1][j+1]) / (n*s-i*j);}}printf("%.4lf\n", dp[0][0]);}return 0;}