poj2096 概率dp入门____期望

最新推荐文章于 2021-03-30 14:32:48 发布

原创最新推荐文章于 2021-03-30 14:32:48 发布 · 445 阅读

1 ·

CC 4.0 BY-SA版权

ACM练习同时被 2 个专栏收录

312 篇文章

订阅专栏

动态规划

88 篇文章

订阅专栏

本文探讨了一种独特的软件Bug收集场景，通过使用动态规划算法解决如何估算完成特定任务所需的平均时间问题。针对一个程序员收集不同类别及子系统的Bug，以达到称软件为“恶心”的标准，文章详细介绍了DP算法的具体实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

首先我要先忏悔一波这学期概率论没好好听课….
哎没想到真的用的上…
我是真的傻逼

期望可以粗略理解为可能的结果*对应的概率
那么本题的dp[a][b] 就可以理解为是a个系统b种bug
因此dp[a][b] 可以从4种状态转移
dp[a+1,b] dp[a,b] dp[a,b+1] dp[a+1,b+1]
首先说一下为什么是从后向前转移
因为后面是已经算好的结果，也就是可能的结果，是已经确定的
因此需要乘以对应概率…
以上四个状态的概率是好求的就不写了..
问题在于他在给自己赋值转移状态的时候用到了他自己…
这个尴尬的问题…
其实也是好解决的..虽然这里的等号是赋值
但是也可以理解为是相等..
那么通过移项来把等式右边的部分给弄掉…
除以系数就求出来了

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<cmath>
#include<cstdio>
#include<queue>
using namespace std;
double dp[1002][1002];
int main()
{
    int n, m;
    cin >> n >> m;
//  dp[n][m] = 1;
    for (int a = n;a >= 0;a--)
    {
        for (int b = m;b >= 0;b--)
        {
            if (a == n&&b == m)continue;
            //dp[a][b] = 1;
            dp[a][b] += (n*m+dp[a][b + 1] * (m - b)*a + dp[a + 1][b] * (n - a)*b + dp[a + 1][b + 1] * (n - a)*(m - b)) / (n*m-a*b);
        }
    }
    printf("%.4f", dp[0][0]);
    return 0;
}