Hiho #1075: 开锁魔法III
Problem Statement
描述
一日,崔克茜来到小马镇表演魔法。
其中有一个节目是开锁咒:舞台上有 n 个盒子,每个盒子中有一把钥匙,对于每个盒子而言有且仅有一把钥匙能打开它。初始时,崔克茜将会随机地选择 k 个盒子用魔法将它们打开。崔克茜想知道最后所有盒子都被打开的概率,你能帮助她回答这个问题吗?
输入
第一行一个整数$T$ ($T \leq 100$)表示数据组数。 对于每组数据,第一行有两个整数$n$和$k$ ($1 \leq n \leq 300, 0 \leq k \leq n$)。 第二行有$n$个整数$a_i$,表示第$i$个盒子中,装有可以打开第$a_i$个盒子的钥匙。
输出
对于每组询问,输出一行表示对应的答案。要求相对误差不超过四位小数。
样例输入
4
5 1
2 5 4 3 1
5 2
2 5 4 3 1
5 3
2 5 4 3 1
5 4
2 5 4 3 1
样例输出
0.000000000
0.600000000
0.900000000
1.000000000
The problem is to compute the probability that use $k$ keys to open the $n$ boxes. In fact we only need to comupte the number of methods that successfully opening $n$ boxes by $k$ choices. Then dividing $C_n^k$ is the final result. So, let's focus on the more refined problem.
First, let's use some notations to express the problem.
Assume the key in box $i$ can open box $a[i]$. Then, the boxes can be opend from box $1$ to $n$ is {$a[1], a[2], ..., a[n]$}.
If we determine open box $i$, then we'll use the key $a[i]$ to open box $a[i]$ which contains the $a[a[i]]$ box key.
So we can assume the keys in the $n$ boxes as a permutation of numbers {$1, 2, ..., n$}. The math model here is just the permutation group in Abstract Algebra.
In order to open all $n$ boxes, we first need to check how many cycles in the permutation. Because the number of keys we need to open all boxes must be greater than or equal to the number of cycles in permutation.
So, if define the number of keys is $k$, and the number of cycles in the $n$-permutation is $m$, the above states $k \geq m$.
Now, we need to design an algorithm to solve the problem. The basical idea is Dynamic Programming (DP).
In general, the hard part of DP is to form a sub-problem. Here, based on the analysis of the permutation above, we'll set the sub-problem by cycles. Because there're $m$ cycles in the $n$-permutation, we'll use $m$ steps to solve the problem.
Define: dp[i][j] = the number of methods that using $j$ keys to solve first $i$ cycles.
Thus the problem can be expressed as computing $dp[m][k]$.
Next, let's construct the recursion. Assume we need to compute $dp[i][j]$.
- In order to solve first $i$ cycles, we can first solve $i-1$ cycles and then the $i^{th}$ cycle.
- If we use $k_i$ keys to solve the $i^{th}$ cycle, we can use only $j-k_i$ keys to solve the first $i-1$ cycles.
- $k_i$ can vary from $1$ to $j$. (Because the initial status may not need key solving, thus $m$ can vary to $j$.) And $k_i$ can't be greater than the size of $i^{th}$ cycle, denoted as $l_i$. (Because every key is belong to one box, so the number of keys we choose can't be greater than the number of boxes in all.)
- For every fixed $k_i$, we just need to multiply the result of first $i-1$ cycles and the result of $i^{th}$ cycle, i.e. $dp[i-1][j-k_i] * C_{l_i}^{k_i}$
(Every $k_i$ keys can solve the $i^{th}$ cycle, so the result of solving $i^{th}$ cycle is $ C_{l_i}^{k_i}$.)
According to above statements, we can get the recursion equation. Here, we use array $comb[n][m]$ to denote the math combination $C_n^m$.
$$dp[i][j] = \sum_{m=1}^{j}(dp[i-1][j-m]*comb[cycle\_i\_length][m])$$
So, the problem is done. But there're two additional problems we need to solve priori.
- First, for efficiency, we can compute the combination numbers before we do the DP algorithm. The computing is also based on DP thinking:
- Compute the combination number by DP, i.e. the simple math equation $C_i^j = C_{i-1}^j + C_{i-1}^{j-1}$. Code is
for(int i = 0; i < 500; ++i)
for(int j = 0; j <= i; ++j)
comb[i][j] =
(0 == i || 0 == j) ? 1 : comb[i-1][j] + comb[i-1][j-1];
- Second, we need to compute the $m$ sizes of cycles in the $n$-permutation:
- Get the cycle information in a $n$-permutation, including number of cycles and the size of every cycle. Here we use array $perm$ to indicate the permutation of $n$ elements.
vector<int> cycles; //store the cycle information
bool used[500];
memset(used, 0, sizeof used); for (int i = 0; i < n; ++i){
if(used[i] == true)
continue; int num = 0;
int idx = i;
while(!used[idx])
{
++num;
used[idx] = true;
idx = perm[idx];
} cycles.push_back(num);
}
Hiho #1075: 开锁魔法III的更多相关文章
- hihocoder 1075 : 开锁魔法III
描述 一日,崔克茜来到小马镇表演魔法. 其中有一个节目是开锁咒:舞台上有 n 个盒子,每个盒子中有一把钥匙,对于每个盒子而言有且仅有一把钥匙能打开它.初始时,崔克茜将会随机地选择 k 个盒子用魔法将它 ...
- #1075 : 开锁魔法III
描述 一日,崔克茜来到小马镇表演魔法. 其中有一个节目是开锁咒:舞台上有 n 个盒子,每个盒子中有一把钥匙,对于每个盒子而言有且仅有一把钥匙能打开它.初始时,崔克茜将会随机地选择 k 个盒子用魔法将它 ...
- HihoCoder 1075 开锁魔法III(概率DP+组合)
描述 一日,崔克茜来到小马镇表演魔法. 其中有一个节目是开锁咒:舞台上有 n 个盒子,每个盒子中有一把钥匙,对于每个盒子而言有且仅有一把钥匙能打开它.初始时,崔克茜将会随机地选择 k 个盒子用魔法将它 ...
- hihoCode 1075 : 开锁魔法III
时间限制:6000ms 单点时限:1000ms 内存限制:256MB 描述 一日,崔克茜来到小马镇表演魔法. 其中有一个节目是开锁咒:舞台上有 n 个盒子,每个盒子中有一把钥匙,对于每个盒子而言有且仅 ...
- hrb——开锁魔法I——————【规律】
解题思路:从1到n的倒数之和. #include<stdio.h> #include<string.h> #include<algorithm> using nam ...
- hihocoder1075【开锁魔法】
hihocoder1075[开锁魔法] 题意是给你一个 \(1-n\) 的置换,求选 \(k\) 个可以遍历所有点的概率. 题目可以换个模型:有 \(n\) 个球,有 \(cnt\) 种不同的颜色,求 ...
- BZOJ 5004: 开锁魔法II 期望 + 组合
Description 题面:www.lydsy.com/JudgeOnline/upload/task.pdf Input Output 一般概率题有两种套路: 满足条件的方案/总方案. 直接求概率 ...
- bzoj5003: 与链 5004: 开锁魔法II 5005:乒乓游戏
www.lydsy.com/JudgeOnline/upload/task.pdf 第一题题意可以转为选一个长度k的序列,每一项二进制的1的位置被下一项包含,且总和为1,考虑每个二进制位的出现位置,可 ...
- 【bzoj5004】开锁魔法II 组合数学+概率dp
题目描述 有 $n$ 个箱子,每个箱子里有且仅有一把钥匙,每个箱子有且仅有一把钥匙可以将其打开.现在随机打开 $m$ 个箱子,求能够将所有箱子打开的概率. 题解 组合数学+概率dp 题目约定了每个点的 ...
随机推荐
- this高级应用 - 域隔离
在js环境中,this有很多指向(window.dom.object等),巧妙的利用this,可以有效的防止变量或方法被外界污染,保证代码健壮性,实例如下. demo: <!DOCTYPE ht ...
- windows 8 update to windows 8.1
可以参考以下几个链接: http://blogs.windows.com/windows/b/appbuilder/archive/2013/07/24/windows-8-to-windows-8- ...
- 【Web】Sublime Text 3 连接sftp/ftp(远程服务器)
在 Win 下常用 Xftp 软件来和远程服务传递文件,但是要是在项目开发的时候频繁的将远程文件拖到本地编辑然后再传回远程服务器,那真是麻烦无比,但是Sublime中SFTP插件,它让这世界美好了许多 ...
- H3 android 系统编译
http://bbs.ickey.cn/group-topic-id-57981.html [Orange Pi PC试用体验]11编译android源码笔记 编译android和编译linux有点类 ...
- 前端之javascript的数据类型1和BOM对象
一 js对象 BOM对象:browser object model浏览器模型对象 window对象:定时器 DOM对象:文档对象模型 js对象:字符串对象,数组对象,日期对象,math对象 new关键 ...
- spring boot 实现mybatis拦截器
spring boot 实现mybatis拦截器 项目是个报表系统,服务端是简单的Java web架构,直接在请求参数里面加了个query id参数,就是mybatis mapper的query id ...
- 821. Shortest Distance to a Character
class Solution { public: vector<int> shortestToChar(string S, char C) { int len=S.length(); ve ...
- 2018.10.23 hdu2476String painter(区间dp)
传送门 一道挺妙的区间dp. 我们先用区间dp求出第一个串为空串时的最小代价. 然后再加入原本的字符更新答案就行了. 代码: #include<bits/stdc++.h> using n ...
- 使用docker 安装 GITLIB
在安装 gitlib 社区版时,配置老不成功,改成使用docker安装 比较顺利,省事. 1外部卷配置 docker 需要配置一些卷在外部,创建一下git的目录 我们创建一个在home下 创建一个gi ...
- boost--asio
1.asio综述 asio的核心类是io_service,它相当于前摄器模式的Proactor角色,在异步模式下发起的I/O操作,需要定义一个用于回调的完成处理函数,当I/O完成时io_service ...