POJ 2096 Collecting Bugs 期望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 .

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

Hint

0<n, s≤1,000

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About

    有 s 个系统， n 种类型 Bug， 每天都会发现一个 Bug，一个 Bug 只属于一种类型和一个系统，且每个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，求找到 s 个系统和 n 种 Bug 的期望。

Solution

    我们可以定义 DP[i][j] 表示发现 i 种 Bug，j 个系统的期望，这里强调一下，期望是一个值，当你的问题确定后，你的期望也就确定了，不要认为它是一个 “变量”, 我打了个引号，你应该知道我的意思。
    让我们想一想 DP 方程：

DP[i][j]=DP[i][j]∗(i / n)∗(j / s) +

DP[i+1][j]∗(1−i / n)∗(j / s) +

DP[i][j+1]∗(i / n)∗(1−j / n)+

DP[i+1][j+1]∗(1−i / n)∗(1−j / s)+1.

我们整理一下 ：
p1=(i / n)∗(j / s)
p2=(1−i / n)∗(j / s)
p3=(i / n)∗(1−j / n)
p4=(1−i / n)∗(1−j / s)

DP[i][j]=DP[i+1][j]∗p2 +DP[i][j+1]∗p3+DP[i+1][j+1]∗p4+11−p1

Code :

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define clr(a) memset(a, 0, sizeof(a))
#define D double
using namespace std;

inline int read() {
    int i = 0, f = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') f = -1; ch = getchar();
    }
    while(isdigit(ch)) {
        i = (i << 3) + (i << 1) + ch - '0'; ch = getchar();
    }
    return i * f;
}

const int MAXN = 1000 + 5;
D dp[MAXN][MAXN];

int main() {
    int n = read(), s = read();
    for(int i = n; i >= 0; --i)
        for(int j = s; j >= 0; --j) {
            if(i == n && j == s) continue;
            D p1 = 1.0 * i / n * j / s,
            p2 = 1.0 * (n - i) / n * j / s,
            p3 = 1.0 * (n -  i) / n * (s - j) / s,
            p4 = 1.0 * i / n * (s - j) / s;
            dp[i][j] = (dp[i + 1][j] * p2 + dp[i + 1][j + 1] * p3 + dp[i][j + 1] * p4 + 1.0) / (D)(1.0 - p1);
        }
    printf("%.4lf", dp[0][0]);
}