zoj 3604 Help Me Escape (记忆化求期望)
来源:互联网 发布:tinygpsce6电子狗软件 编辑:程序博客网 时间:2024/05/30 05:23
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 11 2 3
Sample Output
6.889
Author: WU, Yingxin
Contest: ZOJ Monthly, August 2012
题意:
一只吸血鬼,有n条路给他走,每次他随机走一条路,
每条路有个限制,如果当时这个吸血鬼的攻击力大于
等于某个值,那么就会花费t天逃出去,否则,花费1天
的时间,并且攻击力增加,问他逃出去的期望
学习下记忆化求期望把,这题蛮好
#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <cmath>using namespace std;const int maxn = 2e4 + 5;double dp[maxn], ti[maxn];int n, f, c[105];double dfs(int f){ if(dp[f] > 0) return dp[f]; dp[f] = 0; for(int i = 1; i <= n; i++) { if(f > c[i]) dp[f] += (int)(ti[i])*1.0/n; else dp[f] += (dfs(f+c[i])+1)*1.0/n; //注意这里是先+1再/n,因为他有的可以成功,有的不能成功,不是都是由子状态转移过来的 } return dp[f];}int main(){ while(~scanf("%d%d", &n, &f)) { for(int i = 1; i <= n; i++) scanf("%d", &c[i]); for(int i = 1; i <= n; i++) ti[i] = (1+sqrt(5.0))/2.0 * c[i] * c[i]; memset(dp, 0, sizeof(dp)); double ans = dfs(f); printf("%.3f\n", ans); } return 0;}
- zoj 3604 Help Me Escape (记忆化求期望)
- ZOJ 3640 Help Me Escape(概率dp+记忆化)
- ZOJ 3640 Help Me Escape(概率dp求期望)
- zoj 3640 Help Me Escape 期望DP 简单题 适合记忆化搜索
- ZOJ 3640Help Me Escape (期望dp)
- zoj 3640 Help Me Escape (概率与期望DP)
- ZOJ 3640 Help Me Escape(概率-期望DP+神坑)
- zoj 3640 Help Me Escape(期望)
- Help Me Escape - ZOJ 3640 期望dp
- [ZOJ 3640]Help Me Escape[期望DP]
- zoj 3640 Help Me Escape 记忆化搜索
- zoj 3640 Help Me Escape 概率dp+dfs记忆化
- ZOJ 3640 Help Me Escape 概率DP 记忆化DFS
- zoj 3640 Help Me Escape(概率dp[记忆搜索])
- Help Me Escape ZOJ
- 【解题报告】 ZOJ 3640 Help Me Escape - 期望dp
- 简单概率dp(期望)-zoj-3640-Help Me Escape
- 【ZOJ】3604 Help Me Escape(概率DP)
- shop363总结
- 第1章 HTML5背景知识
- 游戏手柄(JoyStick)编程控制的一个简单代码(Qt)
- 【Java IO模式】Java BIO NIO AIO总结
- 简单利用git部署网站到服务器
- zoj 3604 Help Me Escape (记忆化求期望)
- 选择排序—堆排序
- oracle数据库的修复
- myeclipse 项目中jsp或者js 文件中的错误是没必要处理的,可以忽略
- GYM 100694 E.SuperHyperMarket(set)
- 第一个MYSQL存储过程以及其中遇到的空格问题
- vb.net 教程 1-9 数组3
- RxAndroid和RxJava结合OkGo示例请求网络图片加载到不同ImageView
- ThinkPHP错误