POJ 2096 期望dp

Language:
Collecting Bugs
Time Limit: 10000MS Memory Limit: 64000K
Total Submissions: 6419 Accepted: 3137
Case 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

分析：期望dp不难看出，然后根据套路习惯性从后往前推，至于为什么，我们可以考虑f[i][j]，i表示有i个bug，j个系统的期望，根据期望定义，我们等于所以sigma（可能天数*概率），于是我们需要从前面1到x天转移过来，显然是在扯淡。于是从后往前，f[i][j]表示i表示有i个bug，j个系统离目标还有多少天，则可以看出：
f[i+1][j]　　不属于i种，属于j个系统之一。　　概率为p1=(n-i)/n* j/s。
f[i][j+1]　　属于i种之一，不属于j个系统。 　概率为p2=(s-j)/s* i/n
f[i+1][j+1] 不属于i种，不属于j个系统。 　　 概率为p3=(n-i)/n*(s-j)/s
f[i][j] 　　 属于i种之一，属于j个系统之一。　 概率为p4=i/n* j/s
p1+p2+p3+p4=1，所以是可行的，初始状态为f[n][s]=0；

# include <iostream># include <cstdio># include <cmath># include <list># include <cstring># include <map># include <ctime># include <algorithm># include <queue>using namespace std;typedef long long ll;int read(){    int f=1,i=0;char ch=getchar();    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') {i=(i<<3)+(i<<1)+ch-'0';ch=getchar();}    return f*i;}double f[1010][1010],p[5];int s,n;int main(){    while(scanf("%d%d",&n,&s)!=EOF){        f[n][s]=0;        for(int i=n;i>=0;--i)            for(int j=s;j>=0;--j){                if(i==n&&j==s) continue;                double x,y;x=i,y=j;                p[1]=y/s*x/n;                p[2]=(s-y)/s*x/n;                p[3]=(n-x)/n*y/s;                p[4]=(n-x)/n*(s-y)/s;                f[i][j]=(f[i][j+1]*p[2]+f[i+1][j]*p[3]+f[i+1][j+1]*p[4]+1)/(1-p[1]);            }        printf("%0.4f\n",f[0][0]);    }}