ZOJ - 3640 Help Me Escape

最新推荐文章于 2019-05-10 21:15:47 发布

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最新推荐文章于 2019-05-10 21:15:47 发布

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分类专栏：训练文章标签： dp

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本文探讨了一个数学问题：一个人需要从多个危险的洞穴中逃脱。若其战斗力超过洞穴难度，则按一定公式计算天数；若不足，则提升战斗力再试。文章提供了一段C++代码实现，用于计算成功逃脱的平均天数。

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Background

    If thou doest well, shalt thou not be accepted? and if thou doest not well, sin lieth at the door. And unto thee shall be his desire, and thou shalt rule over him.
    And Cain talked with Abel his brother: and it came to pass, when they were in the field, that Cain rose up against Abel his brother, and slew him.
    And the LORD said unto Cain, Where is Abel thy brother? And he said, I know not: Am I my brother's keeper?
    And he said, What hast thou done? the voice of thy brother's blood crieth unto me from the ground.
    And now art thou cursed from the earth, which hath opened her mouth to receive thy brother's blood from thy hand;
    When thou tillest the ground, it shall not henceforth yield unto thee her strength; a fugitive and a vagabond shalt thou be in the earth.

—— Bible Chapter 4

Now Cain is unexpectedly trapped in a cave with N paths. Due to LORD's punishment, all the paths are zigzag and dangerous. The difficulty of the ith path is ci.

Then we define f as the fighting capacity of Cain. Every day, Cain will be sent to one of the N paths randomly.

Suppose Cain is in front of the ith path. He can successfully take ti days to escape from the cave as long as his fighting capacity f is larger than ci. Otherwise, he has to keep trying day after day. However, if Cain failed to escape, his fighting capacity would increase ci as the result of actual combat. (A kindly reminder: Cain will never died.)

As for ti, we can easily draw a conclusion that ti is closely related to ci. Let's use the following function to describe their relationship:

After D days, Cain finally escapes from the cave. Please output the expectation of D.

Input

The input consists of several cases. In each case, two positive integers N and f (n≤ 100, f ≤ 10000) are given in the first line. The second line includes N positive integers ci (ci ≤ 10000, 1 ≤ i ≤ N)

Output

For each case, you should output the expectation(3 digits after the decimal point).

Sample Input

3 1
1 2 3

Sample Output

6.889

大致就是有一个人要逃出i个洞穴，当战斗力f超过ci时，花费ti天能逃出这个洞口，不然就花一天时间提升si点战斗力，求逃出去的天数期望。

这题我做的迷迷糊糊的，虽然说知道求期望要逆推，但是这道题也不是很简单的从n逆推到1。

设dp[f]是战斗力为f时逃出去的天数期望，那么dp(i) = ( dp(i+c1) + dp(i+c2) + ... + dp(i+cn) ) / n;这是为啥呢？我也不是最清楚，大概是把过每个山洞的期望加起来总和除以n就是最后的期望……为啥呢。

#include <bits/stdc++.h>

using namespace std;

const int maxn=2e4+10;

double dp[maxn];
double c[maxn];
double p=0.5+sqrt(5.0)/2;
int n,f;
double dfs(int tf)
{
    if(dp[tf]>0)
        return dp[tf];
    for(int i=0;i<n;i++)
        if(tf>c[i])
            dp[tf]+=floor(p*c[i]*c[i])/n;
        else
            dp[tf]+=(dfs(tf+c[i])+1)/n;
    return dp[tf];
}
int main()
{
    while(cin>>n>>f)
    {
        for(int i=0;i<n;i++)
            cin>>c[i];
        memset(dp,0,sizeof(dp));
        printf("%.3f\n",dfs(f));
    }
    return 0;
}