POJ 2096 - Collecting Bugs（概率ＤＰ 求期望）

Collecting Bugs

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 2308 Accepted: 1114Case Time Limit: 2000MS Special Judge

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：

一个软件有s个子系统，会产生n种bug。某人一天发现一个bug,这个bug属于一个子系统，属于一个分类    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    问发现n种bug,每个子系统都发现bug的天数的期望。

入门概率DP

dp[i][j] 表示已有i种bug、j个子系统的前提下，要到达目标所需要发现的ｂｕｇ期望数。

因为期望是线性的，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) +...

对于当前状态i,j

再发现一个bug，可能有下面４中情况：

1.发现的这个bug 在 当前已发现的ｉ种bug之中，并且在 当前已发现ｊ个子系统之中。概率 p1 = i*j/(n*s) 数量 dp[i][j]+1

2.发现的这个bug 在 当前已发现的ｉ种bug之中，并且不在 当前已发现ｊ个子系统之中。概率 p1 = i*(s-j)/(n*s) 数量 dp[i][j+1]+1

3.发现的这个bug 不在 当前已发现的ｉ种bug之中，并且在 当前已发现ｊ个子系统之中。概率 p1 = (n-i)*j/(n*s) 数量 dp[i+1][j]+1

4.发现的这个bug 不在 当前已发现的ｉ种bug之中，并且不在 当前已发现ｊ个子系统之中。概率 p1 = (n-i)*(s-j)/(n*s)数量 dp[i+1][j+1]+1

所以：

dp[i][j] = i*j*(dp[i][j]+1)/(n*s) + i*(s-j)*(dp[i][j+1]+1)/(n*s) + (n-i)*j*(dp[i+1][j]+1)/(n*s) + (n-i)*(s-j)*(dp[i+1][j+1]+1)/(n*s)

==>

dp[i][j] = { i*j*dp[i][j] + 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) + 1;

移项化简==>

dp[i][j] = { i*j*dp[i][j] + 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 }/ (n*s-i*j)

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 1000 + 20;double dp[maxn][maxn];int main() {    int n, s;    while(scanf("%d%d", &n, &s) != EOF) {        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-i)*j*dp[i+1][j] + i*(s-j)*dp[i][j+1] + (n-i)*(s-j)*dp[i+1][j+1] + n*s) / (n*s-i*j);            }        }        printf("%.4f\n", dp[0][0]); //注意POJ G++ %f 而不是%lf    }    return 0;}