Maze

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E. Mazetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

A maze is represented by a tree (an undirected graph, where exactly one way exists between each pair of vertices). In the maze the entrance vertex and the exit vertex are chosen with some probability. The exit from the maze is sought by Deep First Search. If there are several possible ways to move, the move is chosen equiprobably. Consider the following pseudo-code:

DFS(x)
    if x == exit vertex then
        finish search
    flag[x] <- TRUE
    random shuffle the vertices' order in V(x) // here all permutations have equal probability to be chosen
    for i <- 1 to length[V] do
        if flag[V[i]] = FALSE then
            count++;
            DFS(y);
    count++;

V(x) is the list vertices adjacent to x. The flag array is initially filled as FALSE. DFS initially starts with a parameter of an entrance vertex. When the search is finished, variable count will contain the number of moves.

Your task is to count the mathematical expectation of the number of moves one has to do to exit the maze.

Input

The first line determines the number of vertices in the graph n (1 ≤ n ≤ 10⁵). The next n - 1 lines contain pairs of integers a_i and b_i, which show the existence of an edge between a_i and b_i vertices (1 ≤ a_i, b_i ≤ n). It is guaranteed that the given graph is a tree.

Next n lines contain pairs of non-negative numbers x_i and y_i, which represent the probability of choosing the i-th vertex as an entrance and exit correspondingly. The probabilities to choose vertex i as an entrance and an exit equal and correspondingly. The sum of all x_i and the sum of all y_i are positive and do not exceed 10⁶.

Output

Print the expectation of the number of moves. The absolute or relative error should not exceed 10^- 9.

ExamplesInputCopy

2
1 2
0 1
1 0

OutputCopy

1.00000000000000000000

InputCopy

3
1 2
1 3
1 0
0 2
0 3

OutputCopy

2.00000000000000000000

InputCopy

7
1 2
1 3
2 4
2 5
3 6
3 7
1 1
1 1
1 1
1 1
1 1
1 1
1 1

OutputCopy

4.04081632653

Note

In the first sample the entrance vertex is always 1 and the exit vertex is always 2.

In the second sample the entrance vertex is always 1 and the exit vertex with the probability of 2/5 will be 2 of with the probability if 3/5 will be 3. The mathematical expectations for the exit vertices 2 and 3 will be equal (symmetrical cases). During the first move one can go to the exit vertex with the probability of 0.5 or to go to a vertex that's not the exit vertex with the probability of 0.5. In the first case the number of moves equals 1, in the second one it equals 3. The total mathematical expectation is counted as 2 / 5 × (1 × 0.5 + 3 × 0.5) + 3 / 5 × (1 × 0.5 + 3 × 0.5)

题解

根据胡渊明的论文，从根节点S到叶节点T的期望步数是n-1。这题T可能没有选到叶节点，但是被选成T的那个节点的子树里面的节点是不会走到的，也就是T可以看成叶节点，只不过期望步数变成了n-1-(siz[T]-1)=n-siz[T]。

那么树上统计，枚举算一下就好了。时间复杂度\(O(n)\)

贴一下别人的代码

const int maxn = 2e5+1e2;
const int maxm = 2*maxn;

int point[maxn],nxt[maxm],to[maxm];
double val[maxn];
int size[maxn];
double st[maxn],ed[maxn];
int n,m;
double sumst,sumed;
int cnt;
double ans;

void addedge(int x,int y)
{
    nxt[++cnt]=point[x];
    to[cnt]=y;
    point[x]=cnt;
}

void dfs(int x,int fa)
{
    size[x]=1;
    val[x]=st[x];
    for (int i=point[x];i;i=nxt[i])
    {
       int p = to[i];
       if (p==fa) continue;
       dfs(p,x);
       size[x]+=size[p];
       val[x]+=val[p];
    }
} 
int main()
{
  n=read();
  for (int i=1;i<n;i++)
  {
    int x=read(),y=read();
    addedge(x,y);
    addedge(y,x);
  }
  
  for (int i=1;i<=n;i++)
  {
     scanf("%lf%lf",&st[i],&ed[i]);
     sumst+=st[i];
     sumed+=ed[i];
  }
  
  for(int i=1;i<=n;i++)
  {
      st[i]=st[i]/sumst;
      ed[i]=ed[i]/sumed;
  }
  dfs(1,0);
  for (int x=1;x<=n;x++)
  {
     for  (int i=point[x];i;i=nxt[i])
     {
        int p =to[i];
        if (size[p]>=size[x])
          ans=ans+1.0*(1.0-val[x])*1.0*(1.0*n-size[x])*ed[x];
        else
          ans=ans+val[p]*1.0*size[p]*ed[x];
     }
  } 
  
  printf("%.12lf",ans);
  return 0;
}