HDU 4035 2011 All network Tournament e probability DP * * * *

It's too much, I won't.

1 /*2 HDU 40353 4 DP asks for the desired question. 5 Test Instructions:6 there were n rooms, connected by the N-1 Tunnel, which actually formed a tree,7 Starting from Node 1, start walking, there are 3 possibilities in each node I:8 1. Be killed, return to Node 1 (probability is Ki)9 2. Find the exit and get out of the maze (probability is EI)Ten 3. And the point is connected with M-edge, random walk a bar One ask: Out of the maze of the number of sides to go to the expected value.  A  - Set E[i] indicates the expectation of the number of sides to go out of the maze at node I. E[1] is the request.  -  the leaf knot: - E[i] = ki*e[1] + ei*0 + (1-ki-ei) * (E[father[i]] + 1); - = ki*e[1] + (1-ki-ei) *e[father[i]] + (1-KI-EI); -  + Non-leaf nodes: (M is the number of sides connected to the node) - E[i] = ki*e[1] + ei*0 + (1-ki-ei)/m* (e[father[i]]+1 +∑ (e[child[i]]+1)); + = ki*e[1] + (1-ki-ei)/m*e[father[i] + (1-ki-ei)/m*∑ (E[child[i]]) + (1-ki-ei); A  at set to each node: e[i] = ai*e[1] + bi*e[father[i] + Ci; -  - for non-leaf node I, set the child node of J for I, then - ∑ (E[child[i]) =∑e[j] - =∑ (aj*e[1] + bj*e[father[j] + Cj) - =∑ (aj*e[1] + bj*e[i] + Cj) in to bring in the above formula - (1-(1-ki-ei)/m*∑bj) *e[i] = (ki+ (1-ki-ei)/m*∑aj) *e[1] + (1-ki-ei)/m*e[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑cj;< /c10> to this will get + Ai = (ki+ (1-ki-ei)/m*∑aj)/(1-(1-ki-ei)/M*∑BJ); - Bi = (1-ki-ei)/M/(1-(1-ki-ei)/M*∑BJ); the Ci = ((1-ki-ei) + (1-ki-ei)/M*∑CJ)/(1-(1-ki-ei)/M*∑BJ); *  $ for the leaf knot.Panax Notoginseng Ai = ki; - Bi = 1-ki-ei; the Ci = 1-ki-ei; +  A start from the leaf node until the a1,b1,c1 is calculated; the  + e[1] = a1*e[1] + b1*0 + C1; - so $ e[1] = C1/(1-A1); $ If the A1 is approaching 1, there is no solution ... -  - */ the#include <stdio.h> -#include <string.h>Wuyi#include <algorithm> the#include <iostream> -#include <math.h> Wu#include <vector> - using namespacestd; About Const intmaxn=10010; $ Const Doubleeps=1e-9;//here 1e-8 will WA. Set to 1e-9 and 1e-10 to - DoubleK[MAXN],E[MAXN]; - DoubleA[MAXN],B[MAXN],C[MAXN]; -  Avector<int>vec[MAXN];//Save Tree +  the BOOLDfsintTintPre//the root node of T is the pre . - { $     intM=vec[t].size ();//the degree of the point 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<m;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) <eps)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); the    //freopen ("OUT.txt", "w", stdout); the     intT; -     intN; the     intu,v; the     intIcase=0; thescanf"%d",&T);94      while(t--) the     { theicase++; thescanf"%d",&n);98          for(intI=1; i<=n;i++) vec[i].clear (); About          for(intI=1; i<n;i++) -         {101scanf"%d%d",&u,&v);102 Vec[u].push_back (v);103 vec[v].push_back (u);104         } the          for(intI=1; i<=n;i++)106         {107scanf"%LF%LF",&k[i],&e[i]);108K[i]/= -;109E[i]/= -; the         }111printf"Case %d:", icase); the         if(Dfs (1,-1) &&fabs (1-a[1]) >EPS)113         { theprintf"%.6lf\n", c[1]/(1-a[1])); the         } the         Elseprintf"impossible\n");117     }118}

HDU 4035 2011 All network Tournament e probability DP * * * *