POJ 2096 Collecting Bugs（0）

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

解题思路：这题题意一开始没看懂，看懂题意之后会发现是一个很基础的概率DP的题目。

首先我们定义状态dp[i][j]表示已经找到i种bug且j个子系统至少含有1个bug仍需要的时间，很显然dp[n][s]=0。

接下来我们考虑状态转移方程，分为下面4种情况：

（1）下一个bug属于这i种且在这个j个子系统中，对应dp[i][j]

（2）下一个bug属于这i种且不在这j个子系统中，对应dp[i][j+1]

（3）下一个bug不属于这i种且在这j个子系统中，对应dp[i+1][j]

（4）下一个bug不属于这i种且不在这j个子系统中，对应dp[i+1][j+1]

因此可以得到状态转移方程为： dp[i][j] = (i/n) * (j/s) * dp[i][j] + (i/n) * ((s-j)/s) * dp[i][j+1] + (n-i)/n * (j/s) * dp[i+1][j] + (n-i)/n * ((s-j)/s) * dp[i+1][j+1] + 1，采用倒推的方法即可，我们最终求解的目标便是dp[0][0]，即当前收集到0种buf且在0个子系统中需要的时间。

#include <ctime>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;const int maxn = 1010;double dp[maxn][maxn];int n, s;int main() {    while(scanf("%d %d", &n, &s) != EOF) {        memset(dp, 0.0, sizeof(dp));        for(int i = n; i >= 0; --i) {            for(int j = s; j >= 0; --j) {                if(i == n && j == s) continue;                double x1 = i*(s-j)*1.0/(n*s)*dp[i][j+1];                double x2 = (n-i)*j*1.0/(n*s)*dp[i+1][j];                double x3 = (n-i)*(s-j)*1.0/(n*s)*dp[i+1][j+1];                dp[i][j] = (x1+x2+x3+1.0)/(1.0-(i*j*1.0)/(n*s));            }        }        printf("%.4lf\n", dp[0][0]);    }    return 0;}