[POJ]2096 Collecting Bugs 期望dp入门

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 6063   Accepted: 2998 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 Northeastern Europe 2004, Northern Subregion

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    期望学的真的烂的不行， 考试做完其他题就看着期望题发呆， 不会做， 虽然昨天碰巧想清楚了一道期望...于是准备开     始死磕期望. 这道题是期望的入门题， 有一个很好的思路是倒着做.

    题意是有n个种类， s个系统， 程序员每天发现一个bug， 这个bug属于一个系统和一个种类. 问发现完n个种类的bug以     及s个系统的bug的期望天数.

    为什么倒着做？ 因为正着做不好做呗... 如果我们设dp[i][j]为已发现i个bug，j个系统bug的期望天数. 我们会发现状     态很难转移. 比如说dp[i][j]由dp[i-1][j]转移过来的话， 你会发现你不知从dp[i-1][j]转移过来的期望可能要几天， 很不     好算.  并且dp[0][0]初始状态并不知道是多少.

    倒着做的题解:
click here.

    注意有个逻辑关系是， 正推dp[i][j]由dp[i-1][j]转移过来是无法确定天数而不可以， 但是倒着来dp[i][j]可以通过一天演变成dp[i][j], dp[i+1][j]...这些状态所以说是正确的.

    TIPS ：注意POJ上面只能是%.4f

#include<stdio.h>
int n, s;
double dp[1001][1001];
int main(){
scanf("%d%d", &n, &s);
for(int i = n; ~i; --i)
for(int j = s; ~j; --j)
if((i != n) || (j != s))
dp[i][j] = (dp[i+1][j] * (n - i) * j + dp[i][j+1] * i * (s - j) + dp[i+1][j+1] * (n-i) * (s-j) + n * s)
/ (n * s - i * j);
printf("%.4f", dp[0][0]);
}