hdu4035 Maze

hdu4035 Maze

​​ Problem Description When 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. Initially, lxhgww is in room 1. Each room has a dangerous trap. When lxhgww step into a room, he has a possibility to be killed and restart from room 1. Every room also has a hidden exit. Each time lxhgww comes to a room, he has chance to find the exit and escape from this maze.

Unfortunately, lxhgww has no idea about the structure of the whole maze. Therefore, he just chooses a tunnel randomly each time. When he is in a room, he has the same possibility to choose any tunnel connecting that room (including the tunnel he used to come to that room). What is the expect number of tunnels he go through before he find the exit?

Input First line is an integer T (T ≤ 30), the number of test cases.

At the beginning of each case is an integer N (2 ≤ N ≤ 10000), indicates the number of rooms in this case.

Then N-1 pairs of integers X, Y (1 ≤ X, Y ≤ N, X ≠ Y) are given, indicate there is a tunnel between room X and room Y.

Finally, N pairs of integers Ki and Ei (0 ≤ Ki, Ei ≤ 100, Ki + Ei ≤ 100, K1 = E1 = 0) are given, indicate the percent of the possibility of been killed and exit in the ith room.

Output For each test case, output one line “Case k: ”. k is the case id, then the expect number of tunnels lxhgww go through before he exit. The answer with relative error less than 0.0001 will get accepted. If it is not possible to escape from the maze, output “impossible”.

Sample Input 3 3 1 2 1 3 0 0 100 0 0 100 3 1 2 2 3 0 0 100 0 0 100 6 1 2 2 3 1 4 4 5 4 6 0 0 20 30 40 30 50 50 70 10 20 60

Sample Output Case 1: 2.000000 Case 2: impossible Case 3: 2.895522 考虑针对叶子节点和非叶子节点分开计算 设E[x]表示x节点走的边数的期望 m为总边数 叶子节点=ki*E[1]+(1-ki-ei)/m*E[fa] +(1-ki-ei) 非叶子节点=ki*E[1]+(1-ki-ei)/m*E[fa]+∑(y in son[x])(1-ki-ei)/m*E[y] +(1-ki-ei) 考虑将式子化简一下 设叶子节点A[x]=k[x] B[x]=1-k[x]-e[x] C[x]=B[x]; 那么叶子节点就是 把E[y]拆开看即可 把fa那一项 变成E[i]然后带入E[x]的式子进行计算 分别化简即可

Ax = (kx+(1-kx-ex)/m * ∑A[y]) / (1 - (1-kx-ex)/m*∑By); Bx = (1-kx-ex)/m / (1 - (1-kx-ex)/m*∑By); Cx = ( (1-kx-ex)+(1-kx-ex)/m*∑Cy ) / (1 - (1-kx-ex)/m*∑By);

#include#include#include#include#include#define eps 1e-9using namespace std;inline char gc(){ static char now[1<<16],*S,*T; if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;} return *S++;}inline int read(){ int x=0,f=1;char ch=gc(); while(!isdigit(ch)) {if (ch=='-') f=-1;ch=gc();} while(isdigit(ch)) x=x*10+ch-'0',ch=gc(); return x*f;}const int N=11000;struct node{ int y,next;}data[N<<1];int h[N],n,num,T,d[N];double k[N],e[N],A[N],B[N],C[N];inline void dfs(int x,int fa){ A[x]=k[x];B[x]=1-k[x]-e[x];C[x]=1-k[x]-e[x]; if (d[x]==1&&x!=1) return;int m=d[x]; double tmp=1,t=(1-k[x]-e[x])/m;B[x]=t; for (int i=h[x];i;i=data[i].next){ int y=data[i].y;if (y==fa) continue; dfs(y,x);tmp-=t*B[y];A[x]+=t*A[y];C[x]+=t*C[y]; }A[x]/=tmp;B[x]/=tmp;C[x]/=tmp;}int main(){ freopen("hdu4035.in","r",stdin); T=read(); for (int owo=1;owo<=T;++owo){ n=read();num=0;memset(h,0,sizeof(h));memset(d,0,sizeof(d)); for (int i=1;i

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