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
题意：师傅又被抓了，师傅现在在一个树里。第一天他在1号节点；对于每一个节点，有三种可能，一是被妖怪杀死ki，二是被徒儿救走ei，三是第二天等概率地走到相邻的一个节点。问师傅被救走的天数的期望，不能被救走输出“impossible”。
思路：这个题存在后续性，举个例子。如果求从s号节点逃出去的期望dp[s]，那么dp[s]和s的子节点和s的父节点有关，而欲求s的子节点时，子节点又和父节点s有关。。。这个时候就需要我们找一个办法来排除后续性。大概就是找一个很牛逼的公式。这个公式本来是和后续性有关，但是公式之间抵消的后续性。
这里借鉴了大佬的博客：
设 E[i]表示在结点i处，要走出迷宫所要走的边数的期望。E[1]即为所求。
叶子结点：
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);
非叶子结点：（m为与结点相连的边数）
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);
设对每个结点：E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;
对于非叶子结点i，设j为i的孩子结点，则
∑(E[child[i]]) = ∑E[j]
= ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
= ∑(Aj*E[1] + Bj*E[i] + Cj)
带入上面的式子得
(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;
由此可得
Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj);
Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj);
Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj);
对于叶子结点
Ai = ki;
Bi = 1 - ki - ei;
Ci = 1 - ki - ei;
从叶子结点开始，直到算出 A1,B1,C1;
E[1] = A1*E[1] + B1*0 + C1;
所以
E[1] = C1 / (1 - A1);
若 A1趋近于1则无解...
AC代码：
#include <bits/stdc++.h>
typedef long long ll;
const int maxx=10010;
const int inf=0x3f3f3f3f;
const double eps=1e-9;
using namespace std;
double k[maxx],e[maxx];
double A[maxx],B[maxx],C[maxx];
vector<int>a[maxx];
bool dfs(int t,int pre)
{
int m=a[t].size();
A[t]=k[t];
B[t]=(1-k[t]-e[t])/m;
C[t]=1-k[t]-e[t];
double tmp=0;
for(int i=0; i<m; i++)
{
int v=a[t][i];
if(v==pre)
continue;
if(!dfs(v,t))
return false;
A[t]+=(1-k[t]-e[t])/m*A[v];
C[t]+=(1-k[t]-e[t])/m*C[v];
tmp+=(1-k[t]-e[t])/m*B[v];
}
if(fabs(tmp-1)<eps)
return false;
A[t]/=(1-tmp);
B[t]/=(1-tmp);
C[t]/=(1-tmp);
return true;
}
int main()
{
int T;
int n;
int u,v;
int iCase=0;
scanf("%d",&T);
while(T--)
{
iCase++;
scanf("%d",&n);
for(int i=1; i<=n; i++)
a[i].clear();
for(int i=1; i<n; i++)
{
scanf("%d%d",&u,&v);
a[u].push_back(v);
a[v].push_back(u);
}
for(int i=1; i<=n; i++)
{
scanf("%lf%lf",&k[i],&e[i]);
k[i]/=100;
e[i]/=100;
}
printf("Case %d: ",iCase);
if(dfs(1,-1) && fabs(1-A[1])>eps)
{
printf("%.6lf\n",C[1]/(1-A[1]));
}
else
printf("impossible\n");
}
return 0;
}