POJ2096--Collecting Bugs--概率DP

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
题意：可以直接把题目理解为有两个桶，第一个桶装n个球，第二个桶装m个球。每次从第一个桶和第二个桶都拿出一个球染成黑色。问将所有球染成黑色需要多少次操作？
思路：E(aA+bB) = aE(A) + bE(B);

           好吧、、、赶脚我概率论白学了！！！
          每一次操作有4种情况，第一个桶拿出的是未染色，第二个染色。第一个染色，第二个未染色。都未染色或者都染色。
          用dp[i][j]表示第一个桶有i个染色了，第二个桶有j个染色了，将全部染色需要的操作数。
          显然有dp[i][j] = p1*dp[i+1][j] + p2*dp[i][j+1] + p3*dp[i+1][j+1] + p4*dp[i][j] + 1;（对应上面4种可能）
          移项就可以得到dp[i][j] 和 其他三项的关系。
          然后反向循环递推即可。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1008
double dp[maxn][maxn];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)==2)
{
dp
[m] = 0;
for(int i = m-1;i >= 0;i--)
dp
[i] = double(m)/(m-i) + dp
[i+1];
for(int i = n-1;i >= 0;i--)
dp[i][m] = double(n)/(n-i) + dp[i+1][m];
for(int i = n-1;i >= 0;i--)
for(int j = m-1;j >= 0;j--)
dp[i][j] = ((n-i)*j*dp[i+1][j] + i*(m-j)*dp[i][j+1] + (n-i)*(m-j)*dp[i+1][j+1] + n*m)/(n*m-i*j);
printf("%.4lf\n",dp[0][0]);
}
return 0;
}