iso 4035

Channel: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4035 There are n rooms connected by N-1 tunnels. In fact, a tree is formed. Starting from node 1, there are three possibilities for each node I: 1. Killed, return to node 1 (probability is Ki) 2. Find the exit and exit the maze (probability is EI)3. There are m edges connected to the vertex and a Random Edge is taken.The expected value of the number of edges to walk out of the maze. Ideas: Set E

Portal: http://acm.hdu.edu.cn/showproblem.php?pid=4035 problem Descriptionwhen Wake up, Lxhgww find himself in a huge maze.The maze consisted by N rooms and tunnels connecting these rooms. Each pair of rooms are connected by one and only one path. Initially, LXHGWW is in the 1. Each is a dangerous trap. When the LXHGWW step into a, he had a possibility to being killed and restart from the 1. Every also has a hidden exit. Each time LXHGWW comes to a, h

])/Du[x]; - } - if(Fabs (tmp-1) return false; -A[x]/= (1-tmp); inB[x]/= (1-tmp); -C[x]/= (1-tmp); to return true; + } - intMain () { the intT,tcase=0; *scanf"%d",T); $ while(t--){Panax Notoginsengtcase++; -tot=0; thememset (First,0,sizeofFirst ); +Memset (Du,0,sizeofdu); Ascanf"%d",n); the for(intI=1; i){ + intx, y; -scanf"%d%d",x,y); $ Add (x, y); $du[x]++; -du[y]++; - } the for(intI=1; i){ -scanf"%LF%LF",k[i],e[i]);WuyiK[i]/= -; e[i]/=

are m edges connected to the vertex and a Random Edge is taken.The expected value of the number of edges to walk out of the maze.Set E [I] to represent the expected number of edges to walk out of the maze at node I. E [1] is the request.Leaf node:E [I] = Ki * E [1] + EI * 0 + (1-ei EI) * (E [Father [I] + 1 );= Ki * E [1] + (1-0000ei) * E [Father [I] + (1-0000ei );Non-leaf node: (m is the number of edges connected to the node)E [I] = Ki * E [1] + EI * 0 + (1-ei EI) /M * (E [Father [I] + 1 + Σ (E

The game looked at the problem, but there was no train of thought. After the game after the total through the 20+, after the game to see the problem-solving report, because the blogger said too concise, and I have never seen this DP seeking mathematical expectations of the problem, so studied for a long time wood has clear. Only search for "DP for expectations" of the topic, starting from a simple start, the eldest brother Kung Fu, finally understand, so write a detailed problem solving report.

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4035 Probability DP on the tree. MazeTime Limit: 2000/1000 MS (Java/others) memory limit: 65768/65768 K (Java/others) total submission (s): 1626 accepted submission (s): 608 Special JudgeProblem descriptionwhen wake up, lxhgww find himself in a huge maze.The maze consisted by N rooms and tunnels connecting these rooms. each pair of rooms is connected by one and only one path. initiall

HDU 4035 maze Realizing the importance of state transfer and simplifying Equations Problem resolved from http://blog.csdn.net/morgan_xww/article/details/6776947 /**DP to find the expected question.Question:There are n rooms connected by n-1 tunnels, which actually forms a tree,Starting from node 1, there are three possibilities for each node I:1. Killed, return to node 1 (probability is Ki)2. Find the exit and exit the maze (probability is EI)3. The

Tags: blog HTTP Io OS AR for SP 2014 on Question: Click the open link. #include HDU 4035 maze probability DP + tree DP

probability of EI escaping from the maze. If they are neither killed nor escaped. You will randomly select a walking edge and finally ask you the expected number of edges to walk out of this maze. Idea: for the expected DP. It is easy to think that status E [I] indicates that it is currently at the I point. Expected number of edges needed to escape the maze. Then, you can write the state transition equation E [I] = Ki * E [1] + 0 * EI + (1-ei-ki) * Σ (E [J] + 1) /EDS [I]. J is the point connect

This question is in three States at each position. Return to the starting point of death: K [I] Locate Exit End E [I] Do not move P [I] K [I] + E [I] + P [I] = 1; Because we only gave n-1 links to connect all of them together, we can naturally regard this figure as a tree structure. Based on Father's Day and leaf nodes Derivation For more information, see: http://blog.csdn.net/morgan_xww/article/details/6776947/ 1 # include HDU 4035 expectation DP

HDU 4035 Maze probability dp + tree dp, hdu4035 Question: Click the open link. #include 200 dynamic planning details Well, I haven't learned this for a long time. I am also confused. I guess I just got started two months ago ......You can see that what he wrote is ineffective and what the optimal sub-structure is. I am also too large ............Dynamic Planning generally solves two types of problems: optimization, maximum value and minimum valu

The main topic: in a tree-shaped maze, the room as a node. There are n rooms, each room has the probability of a trap for KI, there is an exit probability of EI, if both cases are not present (the probability of pi), then can only make a choice to move to the next room (including may go to the previous room). The root node is 1, and when a trap is encountered, it must return to the root node 1 and start again, and when it encounters an exit, go out of the maze. Ask for the desired number of choi

T thea[t]=K[t]; theB[t]= (1-K[T]-E[T])/m; thec[t]=1-k[t]-E[t]; the Doubletmp=0; - for(intI=0; i) in { the intv=Vec[t][i]; the if(V==pre)Continue; About if(!dfs (v,t))return false; thea[t]+= (1-k[t]-e[t])/m*A[v]; thec[t]+= (1-k[t]-e[t])/m*C[v]; thetmp+= (1-k[t]-e[t])/m*B[v]; + } - if(Fabs (tmp-1) return false; theA[t]/= (1-tmp);BayiB[t]/= (1-tmp); theC[t]/= (1-tmp); the return true; - } - intMain () the { the //freopen ("In.txt", "R", stdin); t

http://acm.hdu.edu.cn/showproblem.php?pid=4035Get:1, the first inference is not a tree. In fact, all the feeling figure, both the look at the number of sides = = count-1 is not set2, sometimes, I tell the child to distinguish the tree is still necessary, that is, only when the DFS, when the number of references to a parent node to indicate the number of parameters3, must pay attention to the probability of DP to the accuracy of the very high need to start writing 1e-8,wa several, changed 1e-10 A

non-leaf node is just more $\sum_{j is I child} e[j]$2, because the expectation is originally to seek the limit significance, if the naked such request is obviously recursive infinite, we need to pass the mathematical skill to lose the formulaFor example we can for convenience, set $e[i]$ general formula for $e[i]=a[i]e[1]+b[i]e[Father]+c[i]$ (why take a father unknown?) That's because $i$ 's son needs to know the expectations of his father (ie, $i$), so we can solve the $e[i]$ by bringing them

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