poj 2096 Collecting Bugs - 概率与期望 - 动态规划

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

---

　　题目大意 有n台电脑，每台电脑都有s种bug，每天等概率在一台电脑中发现一个bug，问所有电脑中至少发现了1种bug，并且所有种bug都被发现期望下需要的天数。

　　用 dp[i][j] 表示在i台电脑中发现了bug，有j种bug被发现到最终状态期望要的天数。

　　讨论下面几种情况

　　1)下一次在新的一台电脑上发现了新的一种bug，由 f[i + 1][j + 1] 转移过来

　　2)下一次在新的一台电脑上没有发现新的一种bug，由 f[i + 1][j] 转移过来

　　3)下一次没有在新的一台电脑上发现了新的一种bug，由 f[i][j + 1] 转移过来

　　4)下一次没有在新的一台电脑上没有发现新的一种bug，由 f[i][j] 转移过来

　　因此列得方程

　　整理得到转移方程。

Code

/**
 * poj
 * Problem#2096
 * Accepted
 * Time: 157ms
 * Memory: 8648k
 */
#include <iostream>
#include <cstdio>
using namespace std;

int n, s;
double f[1005][1005];

inline void init() {
    scanf("%d%d", &n, &s);
}

inline void solve() {
    int ns = n * s;
    
    f[n][s] = 0;
    for(int i = n; ~i; i--)
        for(int j = s; ~j; j--)
            if(i < n || j < s)
                f[i][j] = (f[i + 1][j] * (n - i) * j + f[i][j + 1] * i * (s - j) + f[i + 1][j + 1] * (n - i) * (s - j) + ns) / (ns - i * j);
        
    printf("%.4lf", f[0][0]);
}

int main() {
    init();
    solve();
    return 0;
}