Probability and expected dp
Probability
The size of the possibility of an event A occurs, we called the probability of event A, denoted by P (A).
Suppose something there are n possible outcomes of all, the result is like the probability of each event in which m types of cover A, then P (A) = m / n .
Probability of throwing a dice e.g., less than 3 points is 2/6 = 1/3.
If the result of two events A and B, no intersection is covered, then P (A or B occurs) = P (A) + P (B)
Dice or
P (less than 3 points or greater than 4 points) = 2/2 + 6/6 = 2/3
If the results A and B intersect covered
Then P (A or B occurs) = P (A) + P (B) -P (A and B occur simultaneously)
P (less than 3 points or points to an even number) = 2/6 + 3 / 6-1 / 6 = 2/3
B event referred to as "event A does not occur"
Then P (A) + P (B ) = 1, i.e., P (B) =. 1-P (A)
P (not less than 3 points) = 1-2 / 6 = 2 / 3
in the interfering two things, the event a in which a thing, an event B in another thing in
that P (a and B occur simultaneously) = P (a) * P (B)
throw two dice, P (first and second points is less than 3 points is an even number) = (2/6) × (3/6 ) = 1/6
Expect
A variety of events results, remember the size of the result is x, then the expected value of x represents the average size of the result of the event A, denoted by E (x).
The product of E (x) = probability of each outcome and its size.
For example, it points to note throwing a dice X
E (X) =. 1 * (1/6) + 2 * (1/6) +. 3 * (1/6) +. 4 * (1/6) +. 5 * (1/6) + 6 * (1/6) = 7/2
when c is a constant, then:
E (X + c) = E (X) + c, E (c * X) = E * c (X )
E(x+c)=(1+c)*(1/6)*……*(6+c)*(1/6);
The results were mind two events x, y
E (x + y) = E (x) + E (y)
For example: E (+ language scores in math) = E (language scores) + E (math)
As an example may be proved to be correct;
If two events mutually independent , E (x * y) = E (x) * E (y)
E (language scores in math *) = E (language scores) * E (math)
Calculation of probability and expected to have a common computing skills:
If all program events generated are equally probable, then some probability and expected to be converted to a counting problem, and then divided by the total number of programs can be calculated.
The probability of an event such as seeking qualified A, then A is transformed into line with the number of program counting problem; if expectations for the value of the program is translated into all programs of the sum of the value of the counting problem.
Explanation: E (x) = case1 * p1 + case * p2 + case3 * p3 ...... + casen * pn, because it is equal probability problem, p1 = p2 = ...... = pn, so we can then add up the value of all case then * p;
Probability and expectation calculations
A desired probability calculations often use their addition and multiplication rules.
Especially desirable addition rule, in the desired calculation is commonly used. The difference between seeking a desired maximum and minimum values, respectively, find both the expected value can then make a difference.
When multiplication rules, pay attention to whether an event independent of each other
Expected probability may also be calculated by means of equations.
There are four cards that read 0,1,2,3, and each pulled out a back, repeatedly pumping, 0 out of date. Q. expectations extracted number of times.
Set number of extraction is x, then:
X = X * +. 1. 3 /. 4
X =. 4
1 represents the first time necessarily drawn once, and then for a state, there is 1/4 probability pumped 0, then continue to take the number is zero, so 1/4 * 0, there is a probability of 3/4 able to get the rest of the the need to continue pumping, or expect at this time of x, it is 3/4 * x
Almost feel ↑
BZOJ1867 nails and ball
n<=50；
比较简单的概率dp
如果某一个的钉子(x,y)不见了，那么小球垂直下落，就转移到了(x+2,j+1);
设\(f[i][j]\)为小球经过第i行第j列的概率。
$f[1][1]=1 $（即起状态概率为1
\(f[i][j]=f[i-1][j-1] * [(i-1,j-1)有钉子]*1/2 +f[i-1][j] * [(i-1,j)有钉子]*1/2 +f[i-2][j-1] * [(i-2,j-1)没有钉子]\)
\([(i-1,j-1)有钉子]\)可以看做一个bool的0/1数组，有钉子为1，没钉子为0；
至于分数输出，自定义分数数据类型并用gcd化简分数即可。
Bzoj5004 开锁魔法II
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。
100组数据， k<=n<=300。
总方案数C~n~^m^
题目约定了每个点的入度和出度均为1，因此最终的图一定是若干个环。每个环都至少选择一个点即可满足要求。求概率，实际上就是求方案数，最后再除以总方案数即可。
预处理出每个环的点数 c[i] 以及其后缀和 sum[i] 。
设$ f[i][j] \(表示前 i 个环中选出 j 个点，满足最终条件每个环都选的方案数。初始化\) f[0][0]=1$ 。
枚举 i 和 前 i 个环选的点数 j 、第 i 个环选的点数 k
可得\(f[i][j]=\sum\limits_{k=1}^{c[i]}f[i-1][j-k]*C^k_{c[i]}\)
BZOJ5091 摘苹果
在花园中有n棵苹果树以及m条双向道路，每条道路的两端连接着两棵不同的苹果树。假设第i棵苹果树连接着di条道路。小Q将会按照以下方式去
采摘苹果：
1.随机移动到一棵苹果树下，移动到第i棵苹果树下的概率为di/2m，但不在此采摘。
2.重复以下操作k次：等概率随机选择一条与当前苹果树相连的一条道路，移动到另一棵苹果树下，假设当前位于第i棵苹果树下，则他会采摘ai个苹果，多次经过同一棵苹果树下会重复采摘。
请计算小Q期望摘到多少苹果。 n,k<=100000,m<=200000
首先证明这是等概率事件：
设\(f[i][j]\)表示进行了i次操作走到j的概率，易知\(f[0][j]=\frac{dj}{2m}\)（从哪个点开始摘1）
然后考虑转移：第一步从起点走与起点相连的每一条边的概率都是\(\frac{1}{d_j}\)
然后对于整个图来说，第一步走每条边的概率就是第0步（1操作）选择这个点的概率*\(\frac{1}{d_j}\)，也就是\(\frac{d_j}{2m} * \frac{1}{d_j}=\frac{1}{2m}\)
于是\(f[1][j]=\sum\limits_{(u,j)∈e}\frac{1}{2m}=\frac{d_j}{2m}\)
同理：\(f[i][j]=\sum\limits_{(u,j)∈e}\frac{1}{2m}=\frac{d_j}{2m}\)
此时概率相同了。
\(E(x_1+x_2+……+x_n)=\sum\limits_{i=1}^nE(x_i)=\sum\limits_{i=1}^n\sum\limits_{j=1}^kx_i*f[j][i]*a_i\)
（依照定义还是比较好想的）其中xi表示是否在第i棵苹果树下0/1显然0的情况舍弃；
还有一步化简但是并不想写w
BZOJ4832 抵制克苏恩
你有一个英雄和若干奴隶主，对方每次攻击会从你的英雄和奴隶主中随机选一个造成一点伤害。奴隶主受到攻击后，体力为0则死亡，否则若场上奴隶主少于7个，则召唤一个3点血量的奴隶主。
有T局游戏，每局给出初始奴隶主的数量(<=7)和血量(<=3)，给出k，求对方攻击k次后你的英雄受到的总伤害值的期望。
T<=100， k<=50。
设\(f[i][a][b][c]\)表示还要进行i轮攻击，三种血量的奴隶主数量分别为a(血量为1)b(血量2)c(血量3)时，接下来英雄受到的期望总伤害。
设当前共有s=a+b+c个人，那么有1/s的概率打到英雄，a/s的概率打到血量为1的人，b/s的概率打到血量为2的人，c/s的概率打到血量为3的人。
咋的要这样设计状态倒着dp
打到英雄\(f[i-1][a][b][c]+=(f[i][a][b][c]+1)*\frac{1}{s+1};\)
打到a \(f[i-1][a-1][b][c/c+1]+=f[i][a][b][c]*\frac{a}{s+1};\) 此处c考虑奴隶主数是否<7
打到b \(f[i-1][a+1][b-1][c/c+1]+=f[i][a][b][c]*\frac{b}{s+1}\) 此处c考虑奴隶主数是否<7
打到c \(f[i-1][a][b+1][c-1/c]+=f[i][a][b][c]*\frac{c}{s+1}\) 此处c考虑奴隶主数是否<7
NOIP2016 换教室
小A的学校可以视为一个v个点的无向图，他有n门课程要按顺序上课，其中第i门课程要在节点ai进行，但还有一个备选地点bi。
现在小A有m个申请机会，若申请第i门课，那么将有ki的概率使课程搬到bi进行。每门课最多申请一次，而且要在全部申请完成后才知道是否成功，m次机会不必全部用完。他如何申请才能最小化在上课地点间移动的距离的期望值。求该期望值。
v<=300， n,m<=2000
f[i][j][0/1]表示前i个课程申请了j次，且第i个是否申请时的最小期望值。
$f[i][j][0]=Min(f[i-1][j][0]+dis(a[i-1],a[i]) ,
f[i-1][j][1]+k[i-1]dis(b[i-1],a[i])+(1-k[i-1])dis(a[i-1],a[i])) $
\(f[i][j][1]=Min(f[i-1][j-1][0]+dis(a[i-1],b[i])*k[i]+(1-k[i])*dis(a[i-1],a[i]),\\f[i-1][j-1][1]+dis(b[i-1],b[i])*k[i]*k[i-1]+dis(a[i-1]*b[i])*(1-k[i-1])*k[i]\\+dis(b[i-1],a[i])*k[i-1]*(1-k[i])+dis(a[i-1],a[i])*(1-k[i-1])*(1-k[i]));\)
时间复杂度O(v^3+nm)
BZOJ1076 奖励关
有n轮游戏和m种宝物，每种宝物有分数Pi（可以为负），每轮游戏会等概率抛出一种宝物，你可以选择吃或不吃。第i种宝物还有一个限制集合Si，表示只有在Si中的宝物都吃过后，才能吃第i种宝物。
求最优策略下的期望得分。
n<=100， m<=15
设\(f[i][S]\)为还剩下i轮游戏，吃过的宝物集合为S时，接下来能得到的最大期望得分。
==s=--i--
然后同样是倒着搞，最后答案就是当前一轮游戏都没进行，吃过的宝物为0的情况\(f[n][0]\)；初始状态是\(f[0][S]\)其中S是全集；
然后枚举第n-i轮游戏是不是吃了宝物；
如果没吃，显然第n-i轮和第n-i+1轮的结果相同，即\(f[i][S]=f[i-1][S]\)
如果吃了宝物，那么显然第n-i+1轮就多了一个宝物，又因为倒着转移，所以第n-i+1轮的结果要并上第n-i轮吃掉的宝物k，并且加上这个宝物的分数。即\(f[i][S]=f[i-1][S|(i<<l-1)]+p[l]\)
合起来：\(f[i][S]=max\{f[i-1][S],f[i-1][S|(i<<l-1)]+p[l]\};\)
网络题解：
