New Year Santa Network - CodeForces 500 D 树形dp

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New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.

As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.

Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.

It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.

However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.

Input

The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.

Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.

The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.

Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1,1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.

Output

Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.

Sample test(s)
input
32 3 51 3 351 42 21 22 11 1
output
14.000000000012.00000000008.00000000006.00000000004.0000000000
input
61 5 35 3 26 1 71 4 45 2 351 22 13 54 15 2
output
19.600000000018.600000000016.600000000013.600000000012.6000000000

题意：给你一棵树，随机在里面选三个节点，然后分别连接这三个节点，问距离的和的期望是多少。

思路：首先选法有n*(n-1)*(n-2)/6种，每条边设其两边的节点数分别为a和b，那么这条边用到的情况为a*(a-1)*b/2+b*(b-1)*a/2种，而这三个节点间的每条边都需要用两次。另外我用了非递归的形式。

AC代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<vector>using namespace std;typedef long long ll;struct node{    int v,pos;};node A;vector<node> vc[100010];ll len[100010],use[100010],D[100010],M;double D2[100010],W;int p[100010],vis[100010],point[100010],f[100010];int main(){    int T,t,n,m,i,j,k,temp=0,u,v,pos;    ll a,b;    scanf("%d",&n);    for(i=1;i<n;i++)    {        scanf("%d%d%I64d",&u,&v,&len[i]);        A.v=v;A.pos=i;        vc[u].push_back(A);        A.v=u;        vc[v].push_back(A);    }    p[1]=1;temp=1;vis[1]=1;    for(i=1;i<=n;i++)    {        u=p[i];        k=vc[u].size();        for(j=0;j<k;j++)        {            v=vc[u][j].v;            if(vis[v])              continue;            vis[v]=1;            p[++temp]=v;            point[temp]=vc[u][j].pos;            f[temp]=i;        }    }    for(i=n;i>=1;i--)    {        use[i]++;        D[point[i]]=use[i];        use[f[i]]+=use[i];    }    for(i=1;i<n;i++)    {        a=D[i];b=n-D[i];        D[i]=a*(a-1)*b+b*(b-1)*a;    }    M=(ll)n*(n-1)*(n-2)/6;    for(i=1;i<n;i++)       D2[i]=1.0*D[i]/M;    for(i=1;i<n;i++)       W+=D2[i]*len[i];    scanf("%d",&m);    for(i=1;i<=m;i++)    {        scanf("%d%I64d",&pos,&a);        W-=D2[pos]*(len[pos]-a);        len[pos]=a;        printf("%.10f\n",1.0*W);    }}

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